Common derivatives and differentiation techniques
Use this page to revise the following concepts of common derivatives and differentiation techniques:
- Rules for Common Derivatives
- Differentiation Techniques
- Basic Differentiation
- Chain Rule
- Product Rule
- Quotient Rule
Rules for Common Derivatives
The derivatives for any functions like polynomial, power, exponential, logarithmic and circular functions can be easily found by knowing the derivative for their base function.
For example, the derivative of \(\sin{\left(x\right)}\) is \(\cos{\left(x\right)}\). Some of the common rules are listed below.
| Function | Derivative |
|---|---|
| \(x^n\) | \(\dfrac{d}{dx}\left(x^{n}\right) = nx^{n - 1}\) (The power rule) |
| \(x^{\frac{m}{n}}\) | \(\dfrac{d}{dx}\left(x^{\frac{m}{n}}\right) = \dfrac{m}{n}x^{\frac{m - n}{n}}\) |
| \(e^{x}\) | \(\dfrac{d}{dx}\left(e^{x}\right) = e^{x}\) |
| \(\log_{e}(x)\) | \(\dfrac{d}{dx}\left(\log_{e}\left(x\right)\right) = \dfrac{1}{x}\) |
| \(\sin(x)\) | \(\dfrac{d}{dx}\left(\sin\left(x\right)\right) = \cos\left(x\right)\) |
| \(\cos(x)\) | \(\dfrac{d}{dx}\left(\cos\left(x\right)\right) = -\sin\left(x\right)\) |
| \(\tan(x)\) | \(\dfrac{d}{dx}\left(\tan\left(x\right)\right) = \dfrac{1}{\cos^{2}\left(x\right)}\) \(\dfrac{d}{dx}\left(\tan\left(x\right)\right) = \sec^{2}\left(x\right)\) |
Differentiation Techniques
Differentiation techniques are the methods and rules used to find the derivative of a function. These techniques simplify the process of finding derivatives, especially for complex functions.
These are some of the commonly used techniques:
- Basic Differentiation Rules
- Constant Rule
- Power Rule
- Constant Multiple Rule
- Sum/Difference Rule
- Chain Rule
- Product Rule
- Quotient Rule
Basic Differentiation
The derivative of a constant is zero
\[\frac{d}{dx}\left(c\right) = 0\]
Worked Example
Find the derivative of 12
\[\frac{d}{dx}\left(12\right) = 0\]
\[\frac{d}{dx}\left(x^{n}\right) = nx^{n - 1}\]
Worked Example
Find the derivative of \(x^4\)
\[\frac{d}{dx}\left(x^4\right) = 4x^3\]
\[\frac{d}{dx}\left(c\ \cdot\ f\left(x\right)\right) = c\ \cdot\ \frac{d}{dx}\left(f\left(x\right)\right)\]
Worked Example
Find the derivative of \(3x^4\)
\[\frac{d}{dx}\left(3x^{4}\right) = 3\ \cdot\ \frac{d}{dx}\left(x^{4}\right)\]
\[\frac{d}{dx}\left(f\left(x\right) \pm g\left(x\right)\right) = \frac{d}{dx}\left(f\left(x\right)\right) \pm \frac{d}{dx}\left(g\left(x\right)\right)\]
Worked Example
\[\begin{array}{II}f\left(x\right) = 5x^{3} - 7x^{2} +2 \\ g\left(x\right) = 3x^{4} + 32x\end{array}\]
Find \(h^\prime\left(x\right) = f^\prime\left(x\right)x + g^\prime\left(x\right)\)
\[\begin{align}&h^\prime\left(x\right) = \left(15x^{2} - 14z\right) + \left(12x^{3} + 32\right) \\ &h^\prime\left(x\right) = 12x^{3} + 15x^{2} - 14x +32\end{align}\]
Chain Rule
If \(y = f\left(g\left(x\right)\right)\),
\[\frac{dy}{dx} = f^\prime\left(g\left(x\right)\right)\ \cdot\ g^\prime\left(x\right)\]
Alternative notation for chain rule
\[\frac{dy}{dx} = \frac{dy}{du}\ \cdot\ \frac{du}{dx}\]
Here, \(g\left(x\right)\) is the inner function and replaces \(x\) in \(f\left(x\right)\)
Worked Example
Example 1
Find the derivative of \(y = \sin{(7x)}\)
Let \(u = 7x\)
\(y = \sin\left(u\right)\)
\[\begin{align}&\frac{d}{du}\left(\sin\left(u\right)\right) = \cos\left(u\right) \\ &\frac{d}{dx}\left(7x\right) = 7 \\ \ \\ &\frac{dy}{dx} = \frac{dy}{du}\ \cdot\ \frac{du}{dx} \\ &\frac{dy}{dx} = \cos\left(7x\right)\ \cdot\ 7 \\ &\frac{dy}{dx} = 7\cos\left(7x\right)\end{align}\]
Example 2
Find the derivative of \(y = \log_{e}(e^{\sin(x)})\) by applying the chain rule
This function is composed of three nested functions:
- \(v = \sin\left(x\right)\) (the innermost function)
- \(u = e^{v} = e^{\sin\left(x\right)}\) (the middle function)
- \(y = \log_{e}(u)\) the outer most function).
Since the we have three nested functions, we apply the chain rule iteratively, by differentiating \(y\) with respect to \(u\), then \(u\) with respect to \(v\), then \(v\) with respect to \(x\).
\[\begin{align} &\frac{dv}{dx} = &\frac{dy}{du} = \frac{1}{u} \\ &\frac{du}{dv} = e^{v} \\ \frac{d}{dx}\left(\sin\left(x\right)\right) = \cos\left(x\right) \\ \ \\ &\frac{dy}{dx} = \frac{dy}{du}\ \cdot\ \frac{du}{dv}\ \cdot\ \frac{dv}{dx} \\&\frac{dy}{dx} = \frac{1}{u}\ \cdot\ e^{v}\ \cdot\ \cos\left(x\right) \\ &\frac{dy}{dx} = \frac{1}{e^{\sin\left(x\right)}}\ \cdot\ e^{\sin\left(x\right)}\ \cdot\ \cos\left(x\right) \\ &\frac{dy}{dx} = \cos\left(x\right)\end{align}\]
Check:
For Log and Exponential Laws,
\[\begin{align}&\log_{e}\left(e^{\sin\left(x\right)}\right) = \sin\left(x\right)\quad\because\log_{e}\left(e^{f\left(x\right)}\right) = f\left(x\right) \\ &\frac{d}{dx}\left(\sin\left(x\right)\right) = \cos\left(x\right)\end{align}\]
Example 3
Find \(h^\prime\left(x\right)\) when \(h\left(x\right) = f\left(g\left(x\right)\right)\)
- \(f\left(x\right) = 3x^{2}\)
- \(g\left(x\right) = 7e^{2x}\)
- \(h\left(x\right) = f\left(g\left(x\right)\right)\)
- \(f^\prime\left(x\right) = 6x\) (Power rule)
- \(g^\prime\left(x\right) = 7\frac{d}{dx}\left(e^{2x}\right) = 7\ \cdot\ 2e^{2x} = 14e^{2x}\) (Exponential rule)
\[\begin{align}&h^\prime\left(x\right) = f^\prime\left(g\left(x\right)\right)\ \cdot\ g^\prime\left(x\right)\ \textsf{(Chain rule)} \\ &h^\prime\left(x\right) = 6g\left(x\right)\ \cdot\ 14e^{2x} \\ &h^\prime\left(x\right) = 6\left(7e^{2x}\right)\ \cdot\ 14e^{2x}\end{align}\]
Simplifying and combining terms:
\[h^\prime\left(x\right) = 588e^{4x}\]
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If \(h\left(x\right) = g\left(f\left(x\right)\right)\) that the derivative of \(h\left(x\right)\) is \(h^\prime\left(x\right) = g^\prime\left(f\left(x\right)\right)\ \cdot\ f^\prime\left(x\right)\).
The chain rule states that \(\frac{d}{dx}f\left(g\left(x\right)\right) = f^\prime\left(g\left(x\right)\right)\ \cdot\ g^\prime\left(x\right)\)
Therefore, \(\frac{d}{dx}g\left(f\left(x\right)\right) = g^\prime\left(f\left(x\right)\right)\ \cdot\ f^\prime\left(x\right)\).
\[f\left(x\right) = \log_{e}\left(11x\right)\]
By the chain rule, \(\dfrac{dy}{dx} = \dfrac{dy}{du}\ \cdot\ \dfrac{du}{dx}\)
\[\begin{align} &u = 11x \\ y &= \log_{e}\left(u\right) \\ &\dfrac{d}{dx}11x = 11 \\ &\dfrac{d}{du}\log_{e}\left(u\right) = \dfrac{1}{11x} \end{align}\]
\[\begin{align} \dfrac{dy}{dx} &= \dfrac{dy}{du}\ \cdot\ \dfrac{du}{dx} \\ &= \dfrac{1}{11x}\ \cdot\ 11 = \dfrac{1}{x}\end{align}\]
Thus, \(\frac{dy}{dx}\) is \(\frac{1}{x}\), not \(\frac{1}{11x}\)
\[f\left(x\right) = 12\log_{e}\left(\frac{11}{7}x\right)\]
One possible way of calculating \(f^\prime\left(x\right)\) using the change rule \(\dfrac{dy}{dx} = \dfrac{dy}{du}\ \cdot\ \dfrac{du}{dx}\) is
\[\begin{array}{II}y = 12\log_{e}\left(u\right) \\ u = \frac{11}{7}x \\ \frac{du}{dx} = \frac{11}{7}\end{array}\]
From the rules for common derivatives,
\[\frac{d}{dx}\left(\log_{e}\left(x\right)\right) = \frac{1}{x}\]
Therefore,
\[\begin{array}{II}\frac{d}{du}12\log_{e}\left(u\right) = \frac{12}{u} \\ \frac{dy}{dx} = \frac{12}{\frac{11}{7}x}\ \cdot\ \frac{11}{7} = \frac{12}{x}\end{array}\]
This is not the only way of solving this!
We can find the derivative for \(f\left(x\right) = 5\log_{e}\left(\sin\left(7x^{2}\right)\right)\) using the chain rule \(\dfrac{dy}{dx} = \dfrac{dy}{du}\ \cdot\ \dfrac{du}{dx}\)
\[\begin{array}{II}y = 5\log_{e}\left(e\right) \\ u = \sin\left(v\right) \\ v = 7x^{2}\end{array}\]
\[\begin{align}
&\frac{dv}{dx} = 14x \\
&\frac{du}{dv} = \cos\left(7x^{2}\right) \\
&\frac{dy}{du} = \frac{5}{\sin\left(7x^{2}\right)} \\
&\frac{dy}{dx} = \frac{5}{\sin\left(7x^{2}\right)}\ \cdot\ \cos\left(7x^{2}\right)\ \cdot\ 14x \\
&\frac{dy}{dx} = \frac{70x}{\tan\left(7x^{2}\right)}
\end{align}\]
So the derivative of \(5{log}_{e}\left(\sin(7x^2)\right)\) is not \(\dfrac{70}{cos(7x^2)}\).
Product Rule
The product rule is used in calculus to differentiate a function that is the product of two differentiable functions. Specifically, if you have a function \(h\left(x\right) = f\left(x\right)\ \cdot\ g\left(x\right)\), then \(h^\prime\left(x\right) = f\left(x\right)\ \cdot\ g^\prime\left(x\right) + g\left(x\right)\ \cdot\ f^\prime\left(x\right)\)
Notation commonly used for product rule includes
\[\frac{d}{dx}f(x)\ \cdot\ g(x) = f(x)\ \cdot\ g^\prime(x) + g(x)\ \cdot\ f^\prime(x)\]
or
\[\frac{d}{dx}u\ \cdot\ v = u\ \cdot\ \frac{dv}{dx} + v\ \cdot\ \frac{du}{dx}\]
Use the product rule when the function you're differentiating is explicitly the product of two separate functions of \(x\). For example, \(h(x) = 3x^2\ \cdot\ \cos(4x)\).
Worked Example
Example 1
Find the derivative of \(y = 3x^{2}\ \cdot\ \left(\cos\left(4x\right)\right)\)
\[\begin{array}{II}h^\prime\left(x\right) = f^\prime\left(x\right)\ \cdot\ g^\prime\left(x\right) + g\left(x\right)\ \cdot\ f^\prime\left(x\right) \\ f\left(x\right) = 3x^{2}\end{array}\]
\[\begin{align}g\left(x\right) &= \cos\left(4x\right) \\ h^\prime\left(x\right) &= 3x^{2}\ \cdot\ \frac{d}{dx}\left(\cos\left(4x\right)\right) + \cos\left(4x\right)\ \cdot\ \frac{d}{dx}\left(3x^{2}\right) \\ h^\prime\left(x\right) &= 3x^{2}\left(-4\sin\left(4x\right)\right) + \cos\left(4x\right)\ \cdot\ \frac{d}{dx}\left(3x^{2}\right) \\ h^\prime\left(x\right) &= 6x\cos\left(4x\right) - 12x^{2}\sin\left(4x\right) \\ h^\prime\left(x\right) &= 6x\left(\cos\left(4x\right) - 2x\sin\left(4x\right)\right)\end{align}\]
Example 2
Find the derivative of \(= e^{4x}\ \cdot\ \log_{e}\left(7x\right)\)
\[\begin{align}h^\prime\left(x\right) & = f\left(x\right)\ \cdot\ g^\prime\left(x\right) + g\left(x\right)\ \cdot\ f^\prime\left(x\right) \\ h^\prime\left(x\right) &= e^{4x}\ \cdot\ \frac{d}{dx}\left(\log_{e}\left(7x\right)\right) + \log_{e}\left(7x\right)\ \cdot\ \frac{d}{dx}\left(e^{4x}\right) \\ h^\prime\left(x\right) &= e^{4x}\left(\frac{1}{x}\right) + \log_{e}\left(7x\right)\ \cdot\ 4e^{4x} \\ h^\prime\left(x\right) &= e^{4x}\left(\frac{1}{x} + 4\log_{e}\left(7x\right)\right)\end{align}\]
Example 3
Given the \(h\left(x\right) = 7\sqrt{2x}\ \cdot\ \sin\left(\frac{x}{5}\right)\) find \(h^\prime\left(\frac{5\pi}{2}\right)\)
\[\begin{align}h^\prime\left(x\right) &= f\ \cdot\ g^\prime + g\ \cdot\ f^\prime \\ h^\prime\left(x\right) &= 7\sqrt{2x}\ \cdot\ \frac{1}{5}\cos\left(\frac{x}{5}\right) + \sin\left(\frac{x}{5}\right)\ \cdot\ \frac{7\sqrt{2}}{2\ \cdot\ \sqrt{x}} \\ h^\prime\left(x\right) &= \frac{7\sqrt{2x} \cdot \cos\left(\frac{x}{5}\right)}{5} + \frac{7\sqrt{2}\sin\left(\frac{x}{5}\right)}{2 \cdot \sqrt{x}} \\ h^\prime\left(\frac{5\pi}{2}\right) &= \frac{7\sqrt{2 \cdot \frac{5\pi}{2}} \cdot \cos\left(\frac{\frac{5\pi}{2}}{5}\right)}{5} + \frac{7\sqrt{2} \cdot \sin\left(\frac{\frac{5\pi}{2}}{5}\right)}{2 \cdot \sqrt{\frac{5\pi}{2}}} \\ h^\prime\left(\frac{5\pi}{2}\right) &= \frac{7\sqrt{5\pi} \cdot \cos\left(\frac{\pi}{2}\right)}{2} + \frac{7\sqrt{2} \cdot \sin\left(\frac{\pi}{2}\right)}{2 \cdot \sqrt{\frac{5\pi}{2}}} \\ h^\prime\left(\frac{5\pi}{2}\right) &= \frac{7\sqrt{5\pi}\ \cdot\ 0}{5} + \frac{7\sqrt{2}\ \cdot\ 1}{2\ \cdot\ \sqrt{\frac{5\pi}{1}}} \\ h^\prime\left(\frac{5\pi}{2}\right) &= \frac{7}{\sqrt{5\pi}}\end{align}\]
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The product rule states that if \(h\left(x\right) = f\left(x\right)\ \cdot\ g\left(x\right)\), then \(h^\prime\left(x\right) = f\left(x\right)\ \cdot\ g^\prime\left(x\right) +g\left(x\right)\ \cdot\ f^\prime\left(x\right)\).
The order of addition and multiplication don’t matter here, so
\[\begin{align}h\left(x\right) &= f\left(x\right)\ \cdot\ g\left(x\right) = g\left(x\right)\ \cdot\ f\left(x\right) \\ h^\prime\left(x\right) &= f\left(x\right)\ \cdot\ g^\prime\left(x\right) + g\left(x\right)\ \cdot\ f^\prime\left(x\right)\end{align}\]
If \(y = 3x\ \cdot\ \sin\left(3x\right)\), then
\[\begin{array}{II}u = 3x \\ v = \sin\left(3x\right)\end{array}\]
Applying the product rule:
\[\begin{align}&\frac{d}{dx}u\ \cdot\ v = u\ \cdot\ \frac{dv}{dx} + v\ \cdot\ \frac{du}{dx} \\&\frac{d}{dx}3x\ \cdot\ \sin\left(3x\right) = 3x\ \cdot\ \frac{d}{dx} \\ &\frac{dy}{dx} = 3x\ \cdot\ 3\left(\cos\left(3x\right)\right) + \sin\left(3x\right)\ \cdot\ 3 \\ &\frac{dy}{dx} = 3\left(3x\ \cdot\ \left(\cos\left(3x\right)\right) + \sin\left(3x\right)\right)\end{align}\]
Or in its expanded form,
\[\frac{dy}{dx} = 3\sin\left(3x\right) + 9x \cos\left(3x\right)\]
To find the derivative for \(f\left(x\right) = \left(3x\right)5\log_{e}\left(\frac{11x}{2}\right)\) using the product rule:
\[\begin{array}{II}u = 3x \\ v = 5\log_{e}\left(\frac{11x}{2}\right)\end{array}\]
\[\begin{align}&\frac{dy}{dx} = u\ \cdot\ \frac{dv}{dx} + v\ \cdot\ \frac{du}{dx} \\ &\frac{dy}{dx} = 3x\ \cdot\ \frac{d}{dx}\left(5\log_{e}\left(\frac{11x}{2}\right)\right) + 5\log_{e}\left(\frac{11x}{2}\right)\ \cdot\ \frac{d}{dx}\left(3x\right) \\ &\frac{dy}{dx} = 3x\ \cdot\ \frac{5}{x} + 5\log_{e}\left(\frac{11x}{2}\right)\ \cdot\ 3 \\ &\frac{dy}{dx} = 15\log_{e}\left(\frac{11x}{2}\right) + 15\end{align}\]
Therefore, the statement in Q4 is TRUE.
To find the derivative of \(e^{x^{2}}\ \cdot\ \log_{e}\left(x + 1\right)\) using the product rule:
\[\begin{array}{II}u = e^{x^{2}} \\ v = \log_{e}\left(x+ 1\right)\end{array}\]
\[\begin{align}&\frac{dy}{dx} = u\ \cdot\ \frac{dv}{dx} + v\ \cdot\ \frac{du}{dx} \\ &\frac{dy}{dx} = e^{x^{2}}\ \cdot\ \frac{d}{dx}\left(\log_{e}\left(x + 1\right)\right) + \log_{e}\left(x + 1\right)\ \cdot\ \frac{d}{dx}\left(e^{x^{2}}\right) \\ &\frac{dy}{dx} = e^{x^{2}}\ \cdot\ \frac{1}{1 + x} + \log_{e}\left(x + 1\right)\ \cdot\ 2xe^{x^{2}} \\ &\frac{dy}{dx} = e^{x^{2}}\left(\frac{1}{x + 1} + 2x\log_{e}\left(x + 1\right)\right)\end{align}\]
This is NOT \(\dfrac{e^{x^{2}}}{x + 1} + \log_{e}\left(x + 1\right)\) so the statement in that question is FALSE.
To find the derivative of \(h\left(x\right) = x\ \cdot\ \sin\left(\frac{x}{3}\right)\) using the product rule:
\[\begin{array}{II}f\left(x\right) = x \\ g\left(x\right) = \sin\left(\frac{x}{3}\right)\end{array}\]
Product rule states that
\[\begin{align}&h^\prime\left(x\right) = f\left(x\right)\ \cdot\ g^\prime\left(x\right) + g\left(x\right)\ \cdot\ f^\prime\left(x\right) \\ &h^\prime\left(x\right) = x\ \cdot\ \frac{d}{dx}\left(\sin\left(\frac{x}{3}\right)\right) + \sin\left(\frac{x}{3}\right)\ \cdot\ \frac{d}{dx}\left(x\right) \\ &h^\prime\left(x\right) = x\ \cdot\ \frac{1}{3}\left(\cos\left(\frac{x}{3}\right)\right) + \sin\left(\frac{x}{3}\right)\ \cdot\ 1 \\ & h^\prime\left(x\right) = \frac{x}{3}\left(\cos\left(\frac{x}{3}\right)\right) + \sin\left(\frac{x}{3}\right) \\ &h^\prime\left(\pi\right) = \frac{\pi}{3}\left(\cos\left(\frac{\pi}{3}\right)\right) + \sin\left(\frac{\pi}{3}\right) \\ &h^\prime\left(\pi\right) = \frac{\pi}{3}\ \cdot\ \frac{1}{2} + \frac{\sqrt{3}}{2} \\ &h^\prime\left(\pi\right) = \frac{\pi}{6} + \frac{\sqrt{3}}{2}\end{align}\]
Quotient Rule
The quotient rule is a method in calculus for finding the derivative of a function that is the quotient (division) of two differentiable functions.. Specifically, if you have a function
\(f(x) = \dfrac{g(x)}{h(x)}\), then \(f^\prime\left(x\right) = \dfrac{g^\prime\left(x\right)\ \cdot\ h\left(x\right) - g\left(x\right)\ \cdot\ h^\prime\left(x\right)}{\left(h\left(x\right)\right)^{2}}\)
Notations commonly used for the quotient rule include
if \(y = \dfrac{f(x)}{g(x)}\), that
\[\frac{dy}{dx} = \frac{f^\prime\left(x\right)\ \cdot\ g\left(x\right) - f\left(x\right)\ \cdot\ g^\prime\left(x\right)}{\left(g\left(x\right)\right)^{2}}\]
or
\[\frac{dy}{dx} = \frac{f^\prime\left(x\right)\ \cdot\ g\left(x\right) - f\left(x\right)\ \cdot\ \frac{d}{dx}g\left(x\right)}{\left(g\left(x\right)\right)^{2}}\]
or, for \(y = \dfrac{u}{v}\)
\[\frac{dy}{dx} = \frac{v\ \cdot\ \frac{du}{dx} - u\ \cdot\ \frac{dv}{dx}}{v^{2}}\]
Steps to apply the Quotient Rule
- Differentiate the numerator \(g(x)\) to get \(g^\prime(x)\)
- Differentiat the denominator \(h(x)\) to get \(h^\prime(x)\)
- Substitute into the formula: \(\dfrac{g^\prime\left(x\right)\ \cdot\ h\left(x\right) - g\left(x\right)\ \cdot\ h^\prime\left(x\right)}{\left(h\left(x\right)\right)^{2}}\)
Worked Example
Example 1
\(f\left(x\right) = \dfrac{3x^{2}}{2x + 3}\) find \(f^\prime\left(x\right)\)
\[\begin{align}&f^\prime\left(x\right) = \frac{u^\prime\ \cdot\ v- u\ \cdot\ v^\prime}{v^{2}} \\ &f^\prime\left(x\right) = \frac{6x\ \cdot\ \left(2x + 3\right) -2\ \cdot\ 3x^{2}}{\left(2x + 3\right)^{2}} \\ &f^\prime\left(x\right) = \frac{12x^{2} + 18x - 6x^{2}}{\left(2x + 3\right)^{2}} \\ &f^\prime\left(x\right)x= \frac{6x\left(x + 3\right)}{\left(2x + 3\right)^{2}}\end{align}\]
Example 2
\(f\left(x\right) = \dfrac{\sin\left(x\right)}{x^3}\) find \(f^\prime\left(x\right)\)
\[\begin{align}&f^\prime\left(x\right) = \frac{u^\prime\ \cdot\ v - u\ \cdot\ v^\prime}{v^{2}} \\ &f^\prime\left(x\right) = \frac{\cos\left(x\right)\ \cdot\ x^{3} - 3x^{2}\ \cdot\ \sin \left(x\right)}{\left(x^{3}\right)^{2}} \\ & f^\prime\left(x\right) = \frac{x^{3}\cos\left(x\right) - 3x^{2}\ \cdot\ \sin\left(x\right)}{x^{6}} \\ &f^\prime\left(x\right) = \frac{\cos\left(x\right)}{x^{3}} - \frac{3\sin\left(x\right)}{x^{4}}\end{align}\]
Example 3
\(f\left(x\right) = \dfrac{\log_{e}\left(x\right)}{\sqrt{x}}\) find \(f^\prime\left(e\right)\)
\[\begin{align}&f^\prime\left(x\right) = \frac{u^\prime\ \cdot\ v - u\ \cdot\ v^\prime}{v^{2}} \\ &f^\prime\left(x\right) = \frac{\frac{1}{x}\ \cdot\ \sqrt{x} - \log_{e}\left(x\right)\ \cdot\ \frac{1}{2\sqrt{x}}}{\left(\sqrt{x}\right)^{2}} \\ &f^\prime\left(x\right) = \frac{\frac{\sqrt{x}}{x} - \frac{\log_{e}\left(x\right)}{2\sqrt{x}}}{x} \\ &f^\prime\left(x\right) = \frac{2 - \log_{e}\left(x\right)}{2x^{\frac{3}{2}}} \\ &f^\prime\left(e\right) = \frac{2 - \log_{e}\left(e\right)}{2e^{\frac{3}{2}}} \\ &f^\prime\left(e\right) = \frac{1}{2e^{\frac{3}{2}}}\end{align}\]
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If \(y = \frac{u}{v}\) where \(u\) and \(v\) are functions then\(y^\prime = \dfrac{u^\prime\ \cdot\ v - v^\prime\ \cdot\ u}{v^{2}}\)
This is the equivialent to writing
\[\frac{dy}{dx} = \frac{v\ \cdot\ \frac{du}{dx} - u\ \cdot\ \frac{dv}{dx}}{v^{2}}\]
\[y = \frac{\tan(x)}{3x^{2}}\]
\[\begin{align}y^\prime &= \frac{\frac{1}{\cos^{2}(x)} \cdot\ 3x^{2} - \tan(x)\ \cdot\ 6x}{\left(3x^{2}\right)^{2}} \\ &y^\prime = \frac{\frac{1}{\cos^{2}\left(x\right)}}{3x^{2}} - \frac{\tan\left(x\right)\ \cdot\ 6x}{9x^{4}} \\ & y^\prime = \frac{1}{3x^{2}\ \cdot\ \cos^{2}\left(x\right)} - \frac{2\tan\left(x\right)}{3x^{3}}\end{align}\]
Therefore, the statement in the question is TRUE.
Let \(y = \dfrac{x +1}{\log_{e}\left(x\right)}\)
\[\begin{array}{II}y = \frac{u^\prime\ \cdot\ v - v^\prime\ \cdot\ u}{v^{2}} \\ u = x + 1 \\ v = \log_{e}x\end{array}\]
\[\begin{align}&y^\prime = \frac{1\ \cdot\ \left(\log_{e}x\right) - \left(x + 1\right)\ \cdot\ \frac{1}{x}}{\left(\log_{e}x\right)^{2}} \\ &y^\prime = \frac{1}{\log_{e}x} - \frac{x + 1}{x\left(\log_{e}x\right)^{2}}\end{align}\]
\[\begin{array}{II}f\left(x\right) = \frac{x^{3}}{x - 1} \\ g\left(x\right) = x^{3} \\ h\left(x\right) = x - 1\end{array}\]
The quotient rule tells us that
\[\begin{align}&f^\prime\left(x\right) = \frac{g^\prime\left(x\right)\ \cdot\ h\left(x\right) - g\left(x\right)\ \cdot\ h^\prime\left(x\right)}{\left(h\left(x\right)\right)^{2}} \\ &f^\prime\left(x\right) = \frac{3x^{2}\left(x - 1\right) - 1\left(x^{3}\right)}{\left(x - 1\right)^{2}} \\ &f^\prime\left(x\right) = \frac{3x^{3} - 3x^{2} - x^{3}}{\left(x - 1\right)^{2}} \\ &f^\prime\left(x\right) = \frac{x^{2}\left(2x - 3\right)}{\left(x - 1\right)^{2}}\end{align}\]
Substituting \(x = 2\)
\[\begin{align}&f^\prime\left(2\right) = \frac{2^{2}\left(2\ \cdot\ 2 - 3\right)}{\left(2 - 1\right)^{2}} \\ &f^\prime\left(2\right) = \frac{4\left(4 - 3\right)}{1} \\ &f^\prime\left(2\right) = 4\end{align}\]