Summary: Techniques of differentiation

Tools:

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
The derivative of $f$(quantity) is the derivative of $f$, evaluated at that quantity, times the derivative of the quantity.

For instance, if $f(u) = u^{0.5}$, then

$\displaystyle \frac{d}{dx}[u^{0.5}] = 0.5u^{-0.5}\frac{du}{dx}$ The derivative of a quantity raised to the 0.5 is 0.5 times the quantity raised to the -0.5, times the derivative of the quantity.

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

$\displaystyle \frac{d}{dx}[\color{indianred}{(1+x^2)}^3]$	$\displaystyle {}=3\color{indianred}{(1+x^2)}^2\frac{d}{dx}\color{indianred}{(1+x^2)}$	Generalized power rule
	$\displaystyle {}=3(1+x^2)^22x$
	$\displaystyle {}=6x(1+x^2)^2$

$\displaystyle \frac{d}{dx}\left[\frac{2}{\color{indianred}{(x+x^2)}}\right]$	$\displaystyle {}=\frac{d}{dx}2\color{indianred}{(x+x^2)}^{-1}]$	Rewrite in power form
	$\displaystyle {}=-\color{indianred}{(x+x^2)}^{-2}\frac{d}{dx}[\color{indianred}{(x+x^2)}$	Generalized power rule
	$\displaystyle {}=-(x+x^2)^{-2}(1+2x)$
	$\displaystyle {}=-\frac{(1+2x)}{(x+x^2)^{2}}$	Rewrite in rational form.

$\displaystyle \frac{d}{dx}\left[\sqrt{\color{indianred}{x^2}}\right]$	$\displaystyle {}=\frac{1}{2\sqrt{\color{indianred}{x^2}}}\frac{d}{dx}\color{indianred}{(x^2)}$	Generalized rule for square roots
	$\displaystyle {}=\frac{2x}{2\sqrt{x^2}}=\frac{x}{\sqrt{x^2}}$
You can get the same answer by first writing $\color{indianred}{\sqrt{x^2}}$ in power form as $\color{indianred}{(x^2)^{1/2}}$ and then using the generalized power rule.
Notice that $\color{indianred}{\sqrt{x^2} = \|x\|}$, so the result we just obtained tells us that
$\displaystyle \frac{d}{dx}\|x\|$	$\displaystyle {}=\frac{x}{\|x\|}$	Derivative of the absolute value function
Equivalently, $\displaystyle \frac{d}{dx}\|x\|$	$\displaystyle {}=\frac{\|x\|}{x}$	Alternative form (same as above)

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

$\displaystyle \frac{d}{dx}[2 \ln x]$	$\displaystyle {}=2\frac{d}{dx}[\ln x]$	Consant multiple rule
	$\displaystyle {}=2\cdot\dfrac{1}{x}$	Derivative of the natural logarithm
	${} = \dfrac{2}{x}$

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

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

$\displaystyle \frac{d}{dx}[e^{\color{indianred}{x^2+1}}]$	$\displaystyle {}=e^{\color{indianred}{x^2+1}}\frac{d}{dx}[\color{indianred}{x^2+1}]$	Generalized rule
	$\displaystyle {}=e^{x^2+1}(2x)$
	$\displaystyle {}=2xe^{x^2+1}$

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. $\dfrac{(2x-1)^{-3}}{x}$ 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. $e^{(4x-1)(x+2)}$ can be calculated by first calculating the product $(4x-1)(x+2)$, 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. $\ln[(3x^2-1)^5]$ 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

$\displaystyle \frac{d}{dx}[x \sin x]$	$\displaystyle {}=1\cdot \sin x + x\cos x$	Product rule
	$\displaystyle {}=\sin x + x\cos x$

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

$\displaystyle \frac{d}{dx}[\sec(\color{indianred}{x^3})]$	$\displaystyle {}=\sec(\color{indianred}{x^3})\tan(\color{indianred}{x^3})\frac{d}{dx}[\color{indianred}{x^3}]$	Generalized rule
	$\displaystyle {}=\sec(x^3)\tan(x^3)\cdot 3x^2$
	$\displaystyle {}=3x^2\sec(x^3)\tan(x^3)$

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

$xy+y^3x-e^y=4$ First we take the derivative with respect to $x$ (that is, $d/dx$) of both sides:

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: