📝 Generalized chain rule formulas (14 MCQs)

📖 From Calculus • 3. The Derivation • 14 questions available

←Previous Topic:Chain rule for derivatives📝 Generalized chain rule formulasComputer Algebra Systems differentiation using software:Next Topic→

What is Generalized chain rule formulas?

Definition:
The generalized chain rule extends to multiple compositions, stating that for $y = f(u)$ , $u = g(v)$ , $v = h(x)$ , the derivative is $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$ , chaining derivatives through intermediate variables.

Example:
For $y = \sin(e^{x^2})$ , $\frac{dy}{dx} = \cos(e^{x^2}) \cdot e^{x^2} \cdot 2x$ , applying chain rule twice sequentially.

Reason:
This generalization handles deeply nested functions efficiently, crucial for advanced calculus problems in multivariable contexts and implicit differentiation scenarios requiring systematic variable tracking.

Easy

Medium

Hard

📝 All Generalized chain rule formulas MCQs

Q1. Using the generalized derivative formula, what is $\frac{d}{dx}[\sin(3x^{2}+5)]$ ?

\cos(3x^{2}+5)\cdot6x

✅

\cos(3x^{2}+5)\cdot3x

\sin(3x^{2}+5)\cdot6x

-\cos(3x^{2}+5)\cdot6x

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: We set $u=3x^{2}+5$ . Then $du/dx=6x$ . The derivative of $\sin u$ is $\cos u\,du/dx$ . Substituting gives $\cos(3x^{2}+5)\cdot6x$ . This follows directly from the generalized formula, confirming option A.

Q2. Which statement correctly relates the derivative of $u^{r}$ to the derivative of $\sin u$ ?

\frac{d}{dx}[u^{r}] = r u^{r-1}\frac{d}{dx}[\sin u]

\frac{d}{dx}[u^{r}] = r u^{r-1}\cos u\frac{du}{dx}

\frac{d}{dx}[u^{r}] = r u^{r-1}\frac{du}{dx}

and

\frac{d}{dx}[\sin u] = \cos u\frac{du}{dx}

✅

\frac{d}{dx}[u^{r}] = \frac{d}{dx}[\sin u]

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The power rule gives $\frac{d}{dx}[u^{r}] = r u^{r-1}\frac{du}{dx}$ . The chain rule for $\sin u$ gives $\frac{d}{dx}[\sin u] = \cos u\frac{du}{dx}$ . Option C lists both formulas accurately, allowing a direct comparison of how each outer function contributes a factor multiplied by the inner derivative.

Q3. Compare the derivatives of $g(x)=\sqrt{x^{3}+\csc x}$ and $h(x)=\sqrt{x^{3}}$ . Which statement is true?

A.Both derivatives are identical because

\csc x

differentiates to zero.

B.The derivative of

g(x)

includes an extra term

-\csc x\cot x

in the numerator.

C.The derivative of

h(x)

contains

\csc x

while

g(x)

does not.

g(x)

derivative is larger by a factor of

\csc x

. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Using the generalized formula for $\sqrt{u}$ , we get $\frac{1}{2\sqrt{u}}\,du/dx$ . For $g(x)$ , $du/dx = 3x^{2}-\csc x\cot x$ ; for $h(x)$ , $du/dx = 3x^{2}$ . Hence the extra term $-\csc x\cot x$ appears only in the numerator of $g(x)$ ’s derivative, matching option D.

Q4. Which explanation best describes why the derivative of $\sec u$ contains the factor $\tan u$ multiplied by $du/dx$ ?

A.Because

\sec u

is the reciprocal of

\cos u

and differentiating a reciprocal introduces a negative sign.

B.Because the derivative of

\sec u

\sec u\tan u

and the chain rule adds the factor

du/dx

. ✅

C.BECAUSE

\TAN U

IS THE DERIVATIVE OF

\SEC U

WITHOUT ANY CHAIN RULE.

D.Because

\sec u

and

\tan u

are linearly dependent functions.

💡 Difficulty: medium | ✅ Correct: B

Q5. If $u(x)=1+\cos x$ and we compute $\frac{d}{dx}[\sin(\sqrt{u})]$ , which step logically follows after applying the formula for $\sin u$ ?

A.Multiply

\cos(\sqrt{u})

\frac{1}{2\sqrt{u}}

B.Multiply

\cos(\sqrt{u})

\frac{du}{dx}

C.Multiply

\cos(\sqrt{u})

\frac{du}{dx}

and then by

\frac{1}{2\sqrt{u}}

. ✅

D.No further multiplication is needed.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: First, $\frac{d}{dx}[\sin(\sqrt{u})]=\cos(\sqrt{u})\cdot\frac{d}{dx}[\sqrt{u}]$ . The derivative of $\sqrt{u}$ is $\frac{1}{2\sqrt{u}}\,du/dx$ . Therefore the next logical step is to multiply $\cos(\sqrt{u})$ by both $\frac{du}{dx}$ and $\frac{1}{2\sqrt{u}}$ , which is option C.

Q6. Evaluating $f(u)=u^{-8}$ directly versus rewriting as $\frac{1}{u^{8}}$ yields which outcome?

A.Both approaches give identical derivatives. ✅

B.Direct use yields

-8u^{-9}du/dx

; rewriting gives

-8u^{-9}du/dx

plus an extra factor of 2.

C.Direct use gives a wrong sign.

D.Rewriting changes the exponent, LEADING TO A DIFFERENT DERIVATIVE.

💡 Difficulty: medium | ✅ Correct: A

Q7. In $\frac{d}{dx}[(1+x^{5}\cot x)^{-8}]$ , which principle justifies treating the parentheses as a single inner function before applying the power rule?

A.The product rule must be applied first.

B.The chain rule allows any differentiable inner function to be handled as a unit. ✅

C.The quotient rule is required for powers.

D.The function cannot be differentiated because of the cotangent term.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The expression inside the brackets, $u(x)=1+x^{5}\cot x$ , is differentiable. The chain rule (generalized formula) permits us to first differentiate the outer power, treating $u$ as a placeholder, then multiply by $du/dx$ . This systematic approach matches option B.

Q8. If a student omits $du/dx$ when differentiating $\tan u$ , what happens to the derivative of $\tan(3x)$ ?

A.The result will be

\sec^{2}(3x)

instead of

3\sec^{2}(3x)

. ✅

B.The derivative will be zero.

C.The derivative will be

\sec^{2}(3x)\cdot3

D.The sign of the derivative will be reversed.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Correctly, $\frac{d}{dx}[\tan(3x)]=\sec^{2}(3x)\cdot3$ . Forgetting the factor $du/dx$ drops the multiplicative $3$ , leaving $\sec^{2}(3x)$ . This logical omission is exactly described in option A.

Q9. Analyze the derivative of $F(x)=\sec(\sqrt{\omega x})$ where $\omega$ is constant. Which expression correctly represents $F'(x)$ after sequential application of the formulas?

\sec(\sqrt{\omega x})\tan(\sqrt{\omega x})\cdot\frac{\omega}{2\sqrt{\omega x}}

\sec(\sqrt{\omega x})\tan(\sqrt{\omega x})\cdot\frac{\omega}{\sqrt{\omega x}}

\sec(\sqrt{\omega x})\tan(\sqrt{\omega x})\cdot\frac{1}{2\sqrt{\omega x}}

\sec(\sqrt{\omega x})\tan(\sqrt{\omega x})\cdot\frac{\omega}{2\sqrt{x}}

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: First apply the power rule to the outer $\sec$ : derivative is $\sec(\sqrt{\omega x})\tan(\sqrt{\omega x})$ . Then differentiate the inner $\sqrt{\omega x}$ giving $\frac{\omega}{2\sqrt{\omega x}}$ . Multiplying these yields the expression placed in option D.

Q10. Why does the formula for $u^{3/4}$ contain $\frac{3}{4}u^{-1/4}$ rather than $\frac{3}{4}u^{3/4}$ ?

A.Because the exponent is reduced by one when differentiating a power function. ✅

B.Because the formula mistakenly adds the exponent.

C.Because the derivative of a constant exponent is zero.

D.Because the chain rule does not affect the exponent.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The power rule states $\frac{d}{dx}[u^{n}] = n u^{n-1} du/dx$ . Here $n=3/4$ ; subtracting one gives $-1/4$ , so the factor becomes $\frac{3}{4}u^{-1/4}$ . This reduction of the exponent is the essential reason, captured in option A.

Q11. If $u(x)=\ln(\sin x)$ and we compute $\frac{d}{dx}[e^{u}]$ using the generalized formula, which logical step must be taken next?

A.Multiply

e^{u}

\cos x / \sin x

B.Multiply

e^{u}

\frac{d}{dx}[\ln(\sin x)]

. ✅

C.Multiply

e^{u}

\frac{1}{\sin x}

D.No multiplication is needed because the derivative of

e^{u}

is itself.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The derivative of $e^{u}$ is $e^{u}$ times the derivative of $u$ . Hence after recognizing the outer derivative, we must multiply $e^{u}$ by $\frac{d}{dx}[\ln(\sin x)]$ , which equals $\cot x$ . This logical continuation is described in option B.

Q12. Compare the effort of differentiating $K(x)=(x^{2}-x+2)^{3/4}$ directly versus expanding as a series first. Which statement is accurate?

A.Direct application is more efficient and yields the same result as series expansion.

B.Series expansion is faster for this function.

C.Both methods give different derivatives. ✅

D.The generalized formula cannot be applied to fractional exponents.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Using the generalized formula, we differentiate directly: $\frac{3}{4}(x^{2}-x+2)^{-1/4}(2x-1)$ . Expanding into a series would require many terms and then differentiation, which is far less efficient. Both methods, however, produce the same derivative, confirming option C.

Q13. Which statement best synthesizes the role of the generalized derivative formula in simplifying differentiation of composites like $\tan(x^{2}+1)$ and $\sqrt{x^{3}+\csc x}$ ?

A.It eliminates the need for the chain rule entirely.

B.It provides a systematic way to differentiate the outer function first. then multiply by the inner derivative. ✅

C.It replaces the outer function with its integral.

D.IT ONLY WORKS FOR LINEAR INNER FUNCTIONS.

💡 Difficulty: medium | ✅ Correct: B

Q14. According to the generalized derivative table, what is $\frac{d}{dx}[\cos u]$ ?

-\sin u\cdot du/dx

✅

\sin u\cdot du/dx

-\cos u\cdot du/dx

\cos u\cdot du/dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The table lists the derivative of $\cos u$ as $-\sin u$ multiplied by the derivative of the inner function. Thus $\frac{d}{dx}[\cos u]=-\sin u\cdot du/dx$ , which matches option A.

🔗 Related Topics (MCQs)

📝 Tangent line slope definition calculus MCQs 📝 Average vs instantaneous velocity calculus MCQs 📝 Slopes and rates of change calculus MCQs 📝 Derivative definition first principle calculus MCQs 📝 How to find tangent line equation MCQs

←Previous Topic:Chain rule for derivatives📝 Generalized chain rule formulasComputer Algebra Systems differentiation using software:Next Topic→