AP Calculus BC : 3.2 Implicit Differentiation- Exam Style questions with Answer- MCQ

Question

Given that 3x − tan y = 4, what is \(\frac{\mathrm{d} y}{\mathrm{d} x}\)  in terms of y?
A \(\frac{\mathrm{d} y}{\mathrm{d} x}=3\sin ^{2}y\)
B \(\frac{\mathrm{d} y}{\mathrm{d} x}=3\cos ^{2}y\)
C \(\frac{\mathrm{d} y}{\mathrm{d} x}=3\cos y\cot y\)
D \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{1+9y^{2}}\)

Answer/Explanation

 

Question

What is the slope of the line tangent to the curve \(4y^2+xy−2x^2=3\) at the point \((−1,−1)\) ?

A \(−5\)

B \(−\frac{3}{7}\)

C \(\frac{1}{4}\)

D \(\frac{1}{3}\)

Answer/Explanation

Ans:D

 

Question

If \(cos(4x−y)=x+y\) , then \(\frac{dy}{dx}=\)

A \(−1−sin(4x−y)\)

B \(\frac{2+4sin(4x−y)}{sin(4x−y)}\)

C \(\frac{−1}{1+sin(4x−y)}\)

D \(\frac{1+4sin(4x−y)}{−1+sin(4x−y)}\)

Answer/Explanation

Ans:D

The chain rule is the basis for implicit differentiation as well as the differentiation of a composite function.

\(−sin(4x−y)(4−\frac{dy}{dx})=1+\frac{dy}{dx}\)
\(−4sin(4x−y)+sin(4x−y)\frac{dy}{dx}=1+\frac{dy}{dx}\)
\((−1+sin(4x−y))\frac{dy}{dx}=1+4sin(4x−y)\)

Question

                                   

The point (1,3) lies on the curve in the xy-plane given by the equation \(f(x)g(y)=24+x+y\), where f is a differentiable function of x and g is a differentiable function of y. Selected values of f, f′, g, and g′ are given in the table above. What is the value of \(\frac{dy}{dx}\) at the point (1,3) ?

A \(−11\)

B \(4\)

C \(5\)

D \(13\)

Answer/Explanation

Ans: C

The chain rule is the basis for implicit differentiation.

\({f}'(x)g(y)+f(x){g}'(y)\frac{dy}{dx}=1+\frac{dy}{dx}\)
\((−1+f(x){g}'(y))\frac{dy}{dx}=1−{f}'(x)g(y)⇒\frac{dy}{dx}=\frac{1−{f}'(x)g(y)}{−1+f(x){g}'(y)}\)

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