Chain rule
If $f$ is a differentiable function of $u$ and $u$ is a differentiable function of $x$, then the composite $f(u)$ is a differentiable function of $x$, and
$\displaystyle \frac{d}{dx}[f(u)] = f\prime(u)\frac{du}{dx}$
Chain rule in words
For instance, if $f(u) = u^{0.5}$, then
$\displaystyle \frac{d}{dx}[u^{0.5}] = 0.5u^{0.5}\frac{du}{dx}$
Generalized differentiation rules
As illustrated by the following examples, for each function whose derivative we know, we obtain a corresponing "generalized" rule:
Examples: Chain rule
$\displaystyle \frac{d}{dx}[\color{indianred}{(1+x^2)}^3]$  $\displaystyle {}=3\color{indianred}{(1+x^2)}^2\frac{d}{dx}\color{indianred}{(1+x^2)}$ 
 $\displaystyle {}=3(1+x^2)^22x$ 
 $\displaystyle {}=6x(1+x^2)^2$ 

$\displaystyle \frac{d}{dx}\left[\frac{2}{\color{indianred}{(x+x^2)}}\right]$  $\displaystyle {}=\frac{d}{dx}2\color{indianred}{(x+x^2)}^{1}]$ 
 $\displaystyle {}=\color{indianred}{(x+x^2)}^{2}\frac{d}{dx}[\color{indianred}{(x+x^2)}$ 
 $\displaystyle {}=(x+x^2)^{2}(1+2x)$ 
 $\displaystyle {}=\frac{(1+2x)}{(x+x^2)^{2}}$ 

$\displaystyle \frac{d}{dx}\left[\sqrt{\color{indianred}{x^2}}\right]$  $\displaystyle {}=\frac{1}{2\sqrt{\color{indianred}{x^2}}}\frac{d}{dx}\color{indianred}{(x^2)}$ 
 $\displaystyle {}=\frac{2x}{2\sqrt{x^2}}=\frac{x}{\sqrt{x^2}}$ 
$\color{indianred}{\sqrt{x^2}}$ $\color{indianred}{(x^2)^{1/2}}$ 
$\color{indianred}{\sqrt{x^2} = x}$, 
$\displaystyle \frac{d}{dx}x$  $\displaystyle {}=\frac{x}{x}$ 
Equivalently, $\displaystyle \frac{d}{dx}x$  $\displaystyle {}=\frac{x}{x}$ 
Practice:
Derivatives of logarithmic and exponential functions
The following table summarizes the derivatives of logarithmic and exponential functions, as well as their chain rule counterparts (that is, the logarithmic and exponential functions of
a function).
Examples: Derivatives of logarithmic and exponential functions
$\displaystyle \frac{d}{dx}[2 \ln x]$  $\displaystyle {}=2\frac{d}{dx}[\ln x]$ 
 $\displaystyle {}=2\cdot\dfrac{1}{x}$ 
 ${} = \dfrac{2}{x}$ 

$\displaystyle \frac{d}{dx}[\ln(\color{indianred}{2x})]$  $\displaystyle {}=\frac{1}{\color{indianred}{2x}}\frac{d}{dx}[\color{indianred}{2x}]$ 
 $\displaystyle {}=\frac{1}{2x}\cdot 2$ 
 $\displaystyle {}=\frac{1}{x}$ 

$\displaystyle \frac{d}{dx}[\log_3(\color{indianred}{2x+1})]$  $\displaystyle {}=\frac{1}{(\color{indianred}{2x+1})\ln 3}\frac{d}{dx}[\color{indianred}{2x+1}]$ 
 $\displaystyle {}=\frac{1}{(2x+1)\ln 3}\cdot 2$ 
 $\displaystyle {}=\frac{2}{(2x+1)\ln 3}$ 

$\displaystyle \frac{d}{dx}[e^{\color{indianred}{x^2+1}}]$  $\displaystyle {}=e^{\color{indianred}{x^2+1}}\frac{d}{dx}[\color{indianred}{x^2+1}]$ 
 $\displaystyle {}=e^{x^2+1}(2x)$ 
 $\displaystyle {}=2xe^{x^2+1}$ 
Practice:
Practice:
Combining rules for differentiation: Calculation thought experiment
As we saw in
Part 1 of this summary, the
calculation thought experiment (CTE) is a technique to determine whether to treat an algebraic expression as a product, quotient, sum, or difference. Here we extend the CTE to include powers, logarithms, exponentials, sines, cosines, etc. As before, consider the steps you would use in computing its value. If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient, and so on.
Using the calculation thought experiment (CTE) to differentiate a function
If the CTE says, for instance, that the expression is a power of a smaller expression, then apply the generalized power rule as a first step. This will leave you having to differentiate the smaller expression, and you can use the CTE on these, and so on...
Examples: Combining rules for differentiation: Calculation thought experiment
1. $(3x^2 4)^4$ can be calculated by first calculating the expression in parentheses and then raising it to the fourth power. Since the last step is
raising to a power, we can treat the expression as a
power.
2. $\dfrac{(2x1)^{3}}{x}$ can be calculated by first calculating the numerator and denominator separately, and then dividing one by the other. Since the last step is
division, we can treat the expression as a
quotient.
3. $e^{(4x1)(x+2)}$ can be calculated by first calculating the product $(4x1)(x+2)$, and then raising $e$ to the resulting quantity. Since the last step is
raising $e$ to a quantity, we can treat the expression as
$e$ raised to a quantity.
4. $\ln[(3x^21)^5]$ can be calculated by first calculating the expression in brackets, and then taking the logarithm of the result. Since the last step is
taking a logarithm, we can treat the expression as a
logarithm.
Practice:
Derivatives of trigonometric functions
The following table summarizes the derivatives of the six trigonometric functions, as well as their chain rule counterparts (that is, the sine, cosine, etc. of
a function).
Example: Derivatives of trigonometric functions
$\displaystyle \frac{d}{dx}[x \sin x]$  $\displaystyle {}=1\cdot \sin x + x\cos x$ 
 $\displaystyle {}=\sin x + x\cos x$ 

$\displaystyle \frac{d}{dx}[\cos(\color{indianred}{2x^2+1})]$  $\displaystyle {}=\sin(\color{indianred}{2x^2+1})\frac{d}{dx}[\color{indianred}{2x^2+1}]\quad$ 
 $\displaystyle {}=\sin(\color{indianred}{2x^2+1})\cdot 4x$ 
 $\displaystyle {}=4x\sin(2x^2+1)$ 

$\displaystyle \frac{d}{dx}[\sec(\color{indianred}{x^3})]$  $\displaystyle {}=\sec(\color{indianred}{x^3})\tan(\color{indianred}{x^3})\frac{d}{dx}[\color{indianred}{x^3}]$ 
 $\displaystyle {}=\sec(x^3)\tan(x^3)\cdot 3x^2$ 
 $\displaystyle {}=3x^2\sec(x^3)\tan(x^3)$ 
Implicit Functions and Implicit Differentiation
Given an equation in $x$ and $y$, we can think of $y$ as an
implicit function of $x$. We can find $\dfrac{dy}{dx}$ without first solving for $y$ as follows:
 First, take the derivative with respect to $x$ of both sides of the equation (treating $y$ as "a quantity" in the chain rule).
 Next, solve for $\dfrac{dy}{dx}$. This may give $\dfrac{dy}{dx}$ in terms of both $x$ and $y$.
 To evaluate $\dfrac{dy}{dx}$ at a specific value of $x$ (or $y$), first substitute the given value in the original equation relating $x$ and $y$ to obtain a value for the other variable if necessary, and then substitute the values of $x$ and $y$ in the expression for $\dfrac{dy}{dx}$.
Example: Implicit Functions and Implicit Differentiation
Suppose we want to find $\dfrac{dy}{dx}$ given that
First we take the derivative with respect to $x$ (that is, $d/dx$) of both sides:
$\displaystyle \frac{d}{dx}[\color{indianred}{xy} + y^3xe^y] = \frac{d}{dx}[4]$
$\displaystyle \color{indianred}{y+x\frac{dy}{dx}} + 3y^2\frac{dy}{dx} + y^3  e^y\frac{dy}{dx} = 0$
Now we solve for $\dfrac{dy}{dx}$ by bringing together all terms with $\dfrac{dy}{dx}$ on the left and taking the other terms to the right:
$x\frac{dy}{dx}+3y^2\frac{dy}{dx}x  e^y\frac{dy}{dx} = y y^3$
Now factor out $\dfrac{dy}{dx}$ and then solve for it:
$\displaystyle \frac{dy}{dx}\left[x+3xy^2e^y\right] = yy^3$
$\displaystyle \frac{dy}{dx}=\frac{yy^3}{x+3xy^2e^y}$.