Power functions are generally written as $f(x)=kx^n$ where k and n are constants while x is a variable. The derivative of the function is

$f^{\prime}(x)=\frac{d f}{d x}=k n x^{n-1}$

Example
The derivative of $f(x)=8x^3$ is $24x^2$
The polynomial is composed of terms that are power functions, therefore, its derivative is determined by differentiating term by term. If the term is a constant, for instance, h(x) = k where is a constant, then h'(x) = 0 because

$h(x)=kx^0$ and $h^{\prime}(x)=0*kx^{0-1}=0$

Example
The derivative of $f(x) = 5x^4 - 6x^3 + 4x - 10$ is

$f^{\prime}(x) = 5*4x^{4-1} - 6*3x^{3-1} + 4*x^{1-1} - 0 = 20x^3 - 18x^2+4$

If a function f(x) is a product of two functions, g(x) and h(x) that are differentiable, then the derivative of f(x) = g(x) * h(x) is

$f^{\prime}(x)=\frac{d f}{d x}=g^{\prime}(x) * h(x) + g(x) * h^{\prime}(x)$

Also written as

$f(x)=\frac{d f}{d x}=h(x) \frac{d g(x)}{d x}+g(x) \frac{d h(x)}{d x}$

In simple terms, if

$f^{\prime} = g'h + gh'$

If a function f(x) is a quotient of two functions, g(x) and h(x) that are differentiable, then the derivative of is $f(x) = \frac{g(x)}{h(x)}$ is

$f^{\prime}(x)=\frac{d f}{d x}=\frac{g^{\prime}(x) \cdot h(x)-g(x) \cdot h^{\prime}(x)}{h^{2}(x)}$

In simple terms, if

$f=\frac{g}{h} \text { then } f^{\prime}=\frac{g^{\prime} h-g h^{\prime}}{h^{2}}$

$f(x)=\frac{d f}{d x}=h(x) \frac{d g(x)}{d x}+g(x) \frac{d h(x)}{d x}$

If a function f(x) is expressed as$f(x) = [g(x)]^n$, then its derivative is

$\frac{d f}{d x}=n[g(x)]^{n-1} \cdot \frac{d g}{d x}=[g(x)]^{n-1} g^{\prime}(x)$

Example
The derivative of $f(x) = (5x^3 + 2)^4$ is

$\begin{aligned} f^{\prime}(x) &=4\left(5 x^{3}+2\right)^{4-1} \frac{d}{d x}\left(5 x^{3}+2\right) \\ &=4\left(5 x^{3}+2\right)^{3} \cdot 15 x^{2} \\ &=60 x^{2}\left(5 x^{3}+2\right)^{3} \end{aligned}$

The chain rule is a method for differentiation that works for composite functions. It can also be used to differentiate a power function because such function can be expressed as a composite function. If f(x) = g(h(x)) then f=g(h) and h=h(x) hence

$\frac{d f}{d x}=g^{\prime}(h(x)) \cdot h^{\prime}(x)=\frac{d f}{d h} \cdot \frac{d h}{d x}$

Example
Consider $f(x) = (5x^3 + 2)^4$. Let $h(x) = 5x^3+2$ then$f(h)=h^4$ so that $\frac{df}{dh}=4h^3$ and $\frac{dh}{dx}=15x^2$

$\frac{d f}{d x}=\frac{d f}{d h} \cdot \frac{d h}{d x}=4 h^{3} \cdot 15 x^{2}=60 x^{2}\left(5 x^{3}+2\right)$

We provide a table showing the derivatives of trigonometric functions

Example:
Find the derivative of $sin(2x^2+ \pi)$
We use the chain rule: Let $h(x) = 2x^2 + \pi$ then $f(h) = sin h$ . We have h'(x) = 4x and f'(h) = cos h
Therefore,

$\frac{d f}{d x}=\frac{d f}{d h} \cdot \frac{d h}{d x}=\cos h \cdot 4 x=4 x \cos h=4 x \cos 2 x^{2}+\pi$

Example:
Find the derivative of $f(x)=\sqrt{\sec (4 x)}$
Let $$h(x)=\sec (4 x)$$ and $$f(h)=\sqrt{h}=h^{\frac{1}{2}}$$. We have by chain rule $$h^{\prime}(x)=4 \sec (4 x) \tan (4 x)$$ and $$ f^{\prime}(h)=\frac{1}{2} h^{\frac{1}{2}-1}=\frac{1}{2} h^{-\frac{1}{2}}=\frac{1}{2 \sqrt{h}} $$ Thus, $$ \frac{d f}{d x}=\frac{d f}{d h} \cdot \frac{d h}{d x}=\frac{1}{2 \sqrt{h}} \cdot 4 \sec (4 x) \tan (4 x)=\frac{4 \sec (4 x) \tan (4 x)}{2 \sqrt{\sec (4 x)}}=2 \sqrt{\sec (4 x)} \tan (4 x) $$

The derivative of function $f(x) = a^{g(x)}$ is $f'(x) = a^{g(x)}*g(x)ln a$

If a = e = 2.71..., then a = ln e = 1, hence we have derivative of $f(x) = e^{g(x)}$ being $f'(x) = e^{g(x)} * g'(x)$

Example
Find the derivative of $f(x)=e^{x^{2} \tan x}.$
In this case, $g(x)=x^2tan^2(x)$. To get $g^{\prime}(x)$, we use the product rule. $$ g^{\prime}(x)=2 x(\tan x)+\left(x^{2}\right) \sec ^{2} x=2 x \tan x+x^{2} \sec ^{2} x $$ Therefore, $$ f^{\prime}(x)=e^{g(x)} \cdot g^{\prime}(x)=\left(2 x \tan x+x^{2} \sec ^{2} x\right) e^{x^{2} \tan x} $$

Example
Find the derivative of $f(x)=2^{\frac{x}{x^{2}+1}}$
Let $$g(x)=\frac{x}{x^{2}+1}$$, using the quotient rule, $$ g^{\prime}(x)=\frac{1\left(x^{2}+1\right)-2 x(x)}{\left(x^{2}+1\right)^{2}}=\frac{x^{2}+1-2 x^{2}}{\left(x^{2}+1\right)^{2}}=\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}} $$ Therefore, the derivative is $$ f^{\prime}(x)=a^{g(x)} \cdot g^{\prime}(x) \ln a=2^{\frac{x}{x^{2}+1}} \cdot \frac{1-x^{2}}{\left(x^{2}+1\right)^{2}} \cdot \ln 2=\frac{\left(1-x^{2}\right) 2^{\frac{x}{x^{2}+1}} \ln 2}{\left(x^{2}+1\right)^{2}} $$

The derivative of the function $f(x)=log_a(g(x))$ is $f^{\prime}(x)=\frac{g^{\prime}(x)}{g(x) \ln a}.$ If a=e=2.71..., then ln a=ln e=1, hence we have the derivative of $f(x)=log_e(g(x)) = ln g(x)$ being $f^{\prime}(x)=\frac{g^{\prime}(x)}{g(x)}$

Example
Find the derivative of $f(x)=\ln \left(x^{2}+3 x\right)$
Let $g(x)=x^{2}+3 x$, then $f(g)=\ln g$ .Hence, $g^{\prime}(x)=2 x+3$ and $f^{\prime}(g)=\frac{1}{g}$
Therefore,

$f^{\prime}(x)=\frac{g^{\prime}(x)}{g(x)}=\frac{1}{g} \cdot(2 x+3)=\frac{2 x+3}{x^{2}+3 x}$

Example
Find the derivative of $f(x)=\log_{3}(\cos (2 x))$
Let $$g(x)=\cos (2 x)$$, then $$f(g)=\log _{3} g$$. By the chain rule, we have, $$g^{\prime}(x)=-2 \sin (2 x)$$ and $$f^{\prime}(g)=\frac{1}{g \ln 3}$$ Therefore, $$ f^{\prime}(x)=\frac{g^{\prime}(x)}{g(x) \ln a}=\frac{1}{g \ln 3} \cdot(2 \cos (2 x))=\frac{2 \sin (2 x)}{\cos (2 x) \ln 3}=\frac{2 \tan (2 x)}{\ln 3} $$

Constant Multiple Rule

$\frac{d}{dx} (kf) = kf\prime$

Sum (or Difference) Rule

$\frac{d}{dx}(f+g)=f\prime + g\prime \enspace or \enspace \frac{d}{dx}(f-g)=f\prime - g\prime$

Power Rule

$\frac{d}{dx}(x^n) = nx^{n-1}$

Product Rule

$\frac{d}{dx} (fg)=f\prime g + fg\prime$

Quotient Rule

$\frac{d}{dx}(\frac{f}{g})= \frac{f\prime g - fg\prime}{g^2}$

Chain Rule

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Exponential Functions

$\frac{d}{dx}(e^x) = e^x$

Natural Logarithm

$\frac{d}{dx}(\text{ln}x) = \frac{1}{x}$

sine function

$\frac{d}{dx}(sin x) = cos x$

cosine function

$\frac{d}{dx}(cos x) = -sin x$