EduNinja

3.1 The Chain Rule

A composite function has the form $y=f(g(x))$ : the inner function $g$ produces the input for the outer function $f$ . The total derivative multiplies the rate through both stages.

\frac{d}{dx}f(g(x))=f'(g(x))g'(x)\qquad\text{or}\qquad\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

Identify the inner expression $u=g(x)$ .
Differentiate the outer function with respect to $u$ , leaving $u$ in place.
Multiply by $du/dx$ .
Substitute the original inner expression and simplify.

For $y=(3x^2-1)^5$ , let $u=3x^2-1$ . Then $dy/du=5u^4$ and $du/dx=6x$ , so $\frac{dy}{dx}=5(3x^2-1)^4(6x)=30x(3x^2-1)^4.$

Do not stop after differentiating only the outer function: $5(3x^2-1)^4$ is missing the inner derivative $6x$ . The extra factor is unnecessary only when the inner function is exactly $x$ , whose derivative is $1$ .

3.2 Implicit Differentiation

An implicit equation links $x$ and $y$ without necessarily isolating $y$ . Treat $y$ as a differentiable function $y(x)$ . When a term involving $y$ is differentiated with respect to $x$ , the chain rule introduces a factor of $dy/dx$ .

\frac{d}{dx}[F(y)]=F'(y)\frac{dy}{dx}

Differentiate every term on both sides with respect to $x$ .
Attach $dy/dx$ whenever the chain rule differentiates a function of $y$ .
Move all terms containing $dy/dx$ to one side.
Factor out $dy/dx$ and solve for it.

For $x^2+y^2=25$ , differentiate with respect to $x$ : $2x+2y\frac{dy}{dx}=0.$ Therefore $2y\frac{dy}{dx}=-2x\quad\Rightarrow\quad\frac{dy}{dx}=-\frac{x}{y}.$ At $(3,4)$ , the slope is $-3/4$ .

Do not differentiate $y^2$ as merely $2y$ : because $y$ depends on $x$ , the result is $2y\,dy/dx$ . The solved formula $-x/y$ applies where $y\ne 0$ ; when $y=0$ on this circle, the tangent is vertical rather than having a finite $dy/dx$ .

3.3 Differentiating Inverse Functions

If $f(b)=a$ , then $f^{-1}(a)=b$ : the inverse swaps the input and output. Differentiating the identity $f(f^{-1}(x))=x$ with the chain rule shows that the inverse rate is the reciprocal of the original rate at the corresponding point.

\bigl(f^{-1}\bigr)'(a)=\frac{1}{f'\bigl(f^{-1}(a)\bigr)}=\frac{1}{f'(b)},\qquad f(b)=a\text{ and }f'(b)\ne 0

To find $(f^{-1})'(a)$ , locate $b$ such that $f(b)=a$ .
Evaluate the original derivative $f'(b)$ .
Take its reciprocal: $(f^{-1})'(a)=1/f'(b)$ .
An explicit formula for $f^{-1}$ is unnecessary when the corresponding value $b$ is known.

Let $f(x)=x^3+x$ . To find $(f^{-1})'(2)$ , note that $f(1)=2$ , so $f^{-1}(2)=1$ . Since $f'(x)=3x^2+1$ , $\bigl(f^{-1}\bigr)'(2)=\frac{1}{f'(1)}=\frac{1}{3(1)^2+1}=\frac14.$

$f^{-1}(x)$ means the inverse function, not the reciprocal $1/f(x)$ . The reciprocal-rate formula requires an inverse on the relevant interval and $f'(b)\ne 0$ ; if the denominator is zero, this formula does not produce a finite inverse derivative.

3.4 Differentiating Inverse Trigonometric Functions

Inverse trigonometric functions return angles. Their derivative formulas reflect the reciprocal-rate rule for inverse functions. If the input is a differentiable expression $u(x)$ , every formula also includes the inner derivative $u'(x)$ .

\frac{d}{dx}(\arcsin u)=\frac{u'}{\sqrt{1-u^2}},\qquad \frac{d}{dx}(\arccos u)=-\frac{u'}{\sqrt{1-u^2}},\qquad \frac{d}{dx}(\arctan u)=\frac{u'}{1+u^2}

\frac{d}{dx}(\operatorname{arccot}u)=-\frac{u'}{1+u^2},\qquad \frac{d}{dx}(\operatorname{arcsec}u)=\frac{u'}{|u|\sqrt{u^2-1}},\qquad \frac{d}{dx}(\operatorname{arccsc}u)=-\frac{u'}{|u|\sqrt{u^2-1}}

Identify the inverse trigonometric function and its input $u(x)$ .\n2. Write the matching derivative formula.\n3. Substitute $u(x)$ everywhere in the formula.\n4. Multiply the numerator by $u'(x)$ and simplify without losing the required sign.

For $y=\arctan(2x-1)$ , let $u=2x-1$ , so $u'=2$ . Then $\frac{dy}{dx}=\frac{u'}{1+u^2}=\frac{2}{1+(2x-1)^2}.$ The factor $2$ records how quickly the inner expression changes.

$\sin^{-1}x$ means $\arcsin x$ , not $1/\sin x$ . Remember the negative signs for $\arccos$ and $\operatorname{arccot}$ , and the absolute value in the $\operatorname{arcsec}$ and $\operatorname{arccsc}$ denominators. For finite real derivatives, $|u|<1$ in the arcsine/arccosine formulas and $|u|>1$ in the arcsec/arccsc formulas.

3.5 Selecting Procedures for Calculating Derivatives

Choose a derivative procedure from the function's structure before doing algebra. First identify how the largest pieces are connected; then inspect each piece for additional rules.

Sum or difference of terms → differentiate term by term.\n• Product of two changing factors → product rule.\n• Quotient of two changing expressions → quotient rule, unless rewriting makes a simpler equivalent form.\n• One function inside another → chain rule.\n• Equation linking $x$ and $y$ → implicit differentiation.\n• Derivative of an inverse at a corresponding point → inverse-function derivative rule.

Simplify only when it makes the structure clearer.\n2. Mark the outermost operation.\n3. Select the rule for that operation.\n4. Work inward and attach any additional rules, especially the chain rule for composite factors.\n5. Check that every nonconstant factor has been differentiated where the selected rule requires it.

For $y=x^2\sin(3x)$ , the outer structure is a product, so use the product rule. The second factor is composite, so its derivative also needs the chain rule: $\frac{dy}{dx}=2x\sin(3x)+x^2\bigl(3\cos(3x)\bigr).$ Product rule organizes the two factors; chain rule differentiates the inner input $3x$ .

Do not choose a rule from a symbol in isolation. Parentheses may indicate multiplication or composition depending on context. Also, $(uv)'$ is not $u'v'$ and $(f(g(x)))'$ is not just $f'(g(x))$ : the first needs two product-rule terms, while the second needs the inner derivative $g'(x)$ .

3.6 Calculating Higher-Order Derivatives

A higher-order derivative is obtained by differentiating the previous derivative. The second derivative is the derivative of the first derivative; repeating the process produces the third, fourth, and nth derivatives whenever each required derivative exists.

f''(x)=y''=\frac{d^2y}{dx^2},\qquad f^{(n)}(x)=\frac{d^ny}{dx^n}

Differentiate the original function to obtain $f'$ .\n2. Treat $f'$ as the new function and differentiate it to obtain $f''$ .\n3. Continue one order at a time until the requested derivative is reached.\n4. Label each result with its order so that no stage is skipped.

For $f(x)=x^4-2x^3+5$ , $f'(x)=4x^3-6x^2,$ $f''(x)=12x^2-12x,$ and $f'''(x)=24x-12.$ Each line is found by differentiating the line immediately before it.

$d^2y/dx^2$ means differentiate twice; it is not $(dy/dx)^2$ . Also, a requested higher derivative exists only if the preceding derivative is differentiable at the point or throughout the interval being considered.

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

Exam analysis

Most tested topics

3.1 The Chain Rule

Differentiate from the Outside In

3.2 Implicit Differentiation

Differentiate an Implicit Relation

3.3 Differentiating Inverse Functions

Reverse a Function, Reciprocate Its Rate

3.4 Differentiating Inverse Trigonometric Functions

Differentiate Inverse Trig Functions

3.5 Selecting Procedures for Calculating Derivatives

Let Structure Choose the Derivative Rule

3.6 Calculating Higher-Order Derivatives

Differentiate Again for Higher Orders