05 Differentiation Techniques and Related Rates

A progressive guide to derivatives, product and quotient rules, the Chain Rule, implicit and inverse-function methods, higher-order derivatives, and related-rates applications.

Rates of Change and the Basic

A measures how a quantity changes at an instant. For $y=f(x)$ , it is written as $f'(x)=\frac{dy}{dx}$ . Geometrically, $f'(a)$ is the slope of the tangent line at $x=a$ . In applications, derivatives can represent velocity, acceleration, population growth, marginal cost, or changing dimensions.

The basic rules are

\frac{d}{dx}(c)=0,\qquad \frac{d}{dx}(x^n)=nx^{n-1}.

Constant multiples and sums are differentiated term by term:

\frac{d}{dx}[cf(x)]=cf'(x),\qquad \frac{d}{dx}[f(x)\pm g(x)]=f'(x)\pm g'(x).

More complicated expressions require recognizing whether their overall structure is a product, quotient, or composition.

Takeaway: Interpret the first, then identify the structure of the function before choosing a rule.

Products and Quotients

When two differentiable factors are multiplied, use the :

\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x).

Differentiate the first factor while keeping the second fixed, then keep the first factor fixed while differentiating the second.

For

y=(x^2+1)\sin x,

the factors have derivatives $2x$ and $\cos x$ , so

y'=2x\sin x+(x^2+1)\cos x.

The of a product is not generally the product of the derivatives:

(fg)'\ne f'g'.

For a quotient, use the :

\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}.

For

y=\frac{x^2+1}{x-3},

this gives

y'=\frac{(2x)(x-3)-(x^2+1)}{(x-3)^2}=\frac{x^2-6x-1}{(x-3)^2}.

Takeaway: Products produce a sum of two terms, while quotients use a difference in the numerator and the square of the denominator.

Nested Functions and the

A composite function applies one function to the output of another. The is

\frac{d}{dx}f(g(x))=f'(g(x))g'(x).

Differentiate from the outside inward: differentiate the outer function while leaving the inner expression unchanged, then multiply by the of the inner expression.

For

y=(3x^2-5)^4,

the outer function is a fourth power and the inner function is $3x^2-5$ . Therefore,

\frac{dy}{dx}=4(3x^2-5)^3(6x)=24x(3x^2-5)^3.

For several nested layers, repeat the process. If

y=\sin\left((x^2+1)^3\right),

then

y'=\cos\left((x^2+1)^3\right)\cdot 3(x^2+1)^2\cdot 2x.

Thus,

y'=6x(x^2+1)^2\cos\left((x^2+1)^3\right).

Rules can be combined. For

y=(2x+1)^5(3x-2)^7,

use the for the two factors and the within each factor:

y'=10(2x+1)^4(3x-2)^7+21(2x+1)^5(3x-2)^6.

Takeaway: Identify the outer structure first, apply its rule, and then work inward through every composite layer.

is used when an equation relates $x$ and $y$ without isolating $y$ . Treat $y$ as a function of $x$ , so differentiating a term involving $y$ introduces $\frac{dy}{dx}$ .

A reliable procedure is:

Differentiate both sides with respect to $x$ .
Apply the and wherever needed.
Collect all terms containing $\frac{dy}{dx}$ on one side.
Factor out $\frac{dy}{dx}$ and solve.

For the circle

x^2+y^2=25,

differentiation gives

2x+2y\frac{dy}{dx}=0.

Solving produces

\frac{dy}{dx}=-\frac{x}{y}.

At $(3,4)$ , the slope is $-\frac{3}{4}$ .

can also involve products. If

x^3\sin y+y=4x+3,

then

3x^2\sin y+x^3\cos y\frac{dy}{dx}+\frac{dy}{dx}=4.

Collecting terms gives

\left(x^3\cos y+1\right)\frac{dy}{dx}=4-3x^2\sin y.

Therefore,

\frac{dy}{dx}=\frac{4-3x^2\sin y}{x^3\cos y+1}.

Takeaway: Whenever $y$ depends on $x$ , differentiating a term involving $y$ must account for that dependence.

Inverse Functions and Their Derivatives

For a one-to-one differentiable function, an reverses the original input-output relationship. If

f\left(f^{-1}(x)\right)=x,

then differentiating both sides gives

f'\left(f^{-1}(x)\right)(f^{-1})'(x)=1.

Thus,

(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.

At corresponding points, if $f(a)=b$ , then

(f^{-1})'(b)=\frac{1}{f'(a)}.

For example, if $f(2)=5$ and $f'(2)=3$ , then $f^{-1}(5)=2$ , and

(f^{-1})'(5)=\frac{1}{f'(2)}=\frac{1}{3}.

The graph of an inverse is reflected across $y=x$ , which explains why corresponding nonvertical tangent slopes are reciprocals.

For $y=\arcsin x$ , write $\sin y=x$ and differentiate:

\cos y\frac{dy}{dx}=1.

On the range of $\arcsin x$ , $\cos y=\sqrt{1-x^2}$ , so

\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}},\qquad -1<x<1.

Takeaway: Locate the corresponding point on the original function and take the reciprocal of the original .

Repeated Differentiation

are obtained by differentiating a repeatedly. The second is

f''(x)=\frac{d}{dx}[f'(x)]=\frac{d^2y}{dx^2}.

Further derivatives include $f'''(x)$ , $f^{(4)}(x)$ , and $f^{(n)}(x)$ . If $s(t)$ is position, then

v(t)=s'(t)

is velocity and

a(t)=s''(t)

is acceleration.

For

f(x)=x^5-3x^3+2x,

successive differentiation gives

f'(x)=5x^4-9x^2+2,

f''(x)=20x^3-18x,

f'''(x)=60x^2-18,

f^{(4)}(x)=120x,

and

f^{(5)}(x)=120.

The second also describes how a function's rate of change is itself changing, including concavity in appropriate contexts.

Takeaway: Repeated differentiation reveals successive layers of change, from position to velocity to acceleration and beyond.

in Applications

problems connect quantities that change with time through an equation. The central strategy is to differentiate that relationship with respect to time before inserting numerical values.

Use this sequence:

Name the changing quantities.
State the known and unknown rates with units.
Write an equation relating the quantities.
Differentiate with respect to time.
Substitute the specified values.
Solve for the unknown rate and interpret its sign and units.

For a spherical balloon,

V=\frac{4}{3}\pi r^3.

Differentiating with respect to time gives

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.

If $\frac{dV}{dt}=2\text{ cm}^3/\text{s}$ when $r=3\text{ cm}$ , then

2=4\pi(3)^2\frac{dr}{dt},

\frac{dr}{dt}=\frac{1}{18\pi}\text{ cm/s}.

The positive sign means the radius is increasing.

For a $10$ -ft ladder, let $x$ be the bottom's distance from the wall and $y$ the top's height. The relationship is

x^2+y^2=100.

If $\frac{dx}{dt}=2\text{ ft/s}$ , differentiate:

2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.

When $x=6$ , the Pythagorean relationship gives $y=8$ . Therefore,

2(6)(2)+2(8)\frac{dy}{dt}=0,

\frac{dy}{dt}=-\frac{3}{2}\text{ ft/s}.

The negative sign means the top is moving downward.

Takeaway: Differentiate first, substitute second, and use the sign and units to explain the result.

Strategy and Error Checks

Several mistakes can be prevented by checking the structure of the expression and the meaning of the result.

For a product, use $(fg)'=f'g+fg'$ , not $f'g'$ .
For a composition, use $(f\circ g)'=f'(g(x))g'(x)$ , including the of the inner function.
In implicit work, remember that $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ , not merely $2y$ .
In related-rates problems, do not substitute a numerical value before differentiating; doing so can remove the changing quantity whose rate is needed.
Include units for rates, such as $\text{cm/s}$ , $\text{ft/s}$ , or square units per second for an area rate.
Check whether a negative answer indicates decrease or downward motion, rather than treating the sign as an error.

A useful final check is to ask whether the rule matched the structure, whether every changing variable was differentiated correctly, and whether the units match the requested rate.

Final takeaway: Correct differentiation depends on both symbolic technique and interpretation. Recognize the structure, preserve dependencies, delay substitution, and interpret the result.

Rates of Change and the Basic Derivative