📝 Implicit differentiation differentiability (14 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 14 questions available

←Previous Topic:Implicit differentiation examples📝 Implicit differentiation differentiabilityDerivatives of Logarithmic Functions:Next Topic→

What is Implicit differentiation differentiability?

Definition:
A curve defined implicitly by $F(x, y) = 0$ is differentiable at a point if the partial derivatives $F_x$ and $F_y$ exist and are continuous, and specifically if $F_y \neq 0$ at that point. This condition ensures that the Implicit Function Theorem applies, guaranteeing that $y$ can be locally expressed as a differentiable function of $x$ near that specific point.

Example:
For $x^2 + y^2 = 1$ , $F_y = 2y$ . At $(0, 1)$ , $F_y = 2 \neq 0$ , so it is differentiable. However, at $(1, 0)$ , $F_y = 0$ , indicating a vertical tangent where the derivative is undefined, showing non-differentiability in the standard sense.

Reason:
Understanding differentiability conditions prevents errors when applying implicit differentiation, ensuring that the resulting derivative represents a valid tangent line and that the function behaves smoothly without vertical tangents or cusps at the point of interest.

Easy

Medium

Hard

📝 All Implicit differentiation differentiability MCQs

Q1. For an equation $F(x,y)=0$ to define $y$ as a differentiable function of $x$ near a point $(x_0,y_0)$ , which condition must be satisfied?

F

is continuous at

(x_0,y_0)

\frac{\partial F}{\partial y}(x_0,y_0)\neq 0

✅

\frac{\partial F}{\partial x}(x_0,y_0)=0

F(x_0,y_0)=0

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The Implicit Function Theorem guarantees a local function $y(x)$ when the partial derivative with respect to $y$ is non‑zero at the point. Continuity alone does not ensure solvability, and the equation itself being zero is already assumed. Hence the non‑vanishing of $\partial F/\partial y$ is the essential condition.

Q2. Why does implicit differentiation of the equation $x^{2}+y^{2}+1=0$ produce a meaningless derivative?

A.Because the differentiation formula was applied incorrectly

B.Because there are no real points satisfying the equation ✅

C.Because

\frac{\partial F}{\partial y}=0

everywhere

D.Because

y

is not a function of

x

in any neighbourhood

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The equation $x^{2}+y^{2}+1=0$ has no real solutions, so no real curve exists on which a function $y(x)$ could be defined. Differentiating an equation that never holds yields a formal expression for $dy/dx$ that has no geometric meaning, illustrating the need to verify existence of a real implicit function first.

Q3. The equation $xy+2y=1$ can be solved for $y$ as $y=\frac{1}{x+2}$ . What is $\frac{dy}{dx}$ ?

-\frac{1}{(x+2)^{2}}

✅

-\frac{1}{(x+2)^{2}}

-\frac{1}{x+2}

\frac{1}{(x+2)^{2}}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Solving for $y$ gives $y=(x+2)^{-1}$ . Differentiating directly, $\frac{dy}{dx}=-1\cdot (x+2)^{-2}$ , which simplifies to $-\frac{1}{(x+2)^{2}}$ . Implicit differentiation of the original equation leads to the same expression, confirming consistency between the two methods.

Q4. For the implicit relation $\sin y = x$ , what is the set of $x$ values for which a differentiable function $y(x)$ can be defined?

-1\le x\le 1

✅

B.All real numbers

x>0

x<0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The sine function ranges only between $-1$ and $1$ . Therefore, real solutions $y$ exist only when $-1\le x\le 1$ . Within this interval the inverse function $\arcsin x$ is differentiable, so the implicit definition yields a differentiable function of $x$ precisely on that domain.

Q5. Using implicit differentiation on $x^{2}-y^{3}=xy$ at the point $(1,1)$ , what is $\frac{dy}{dx}$ ?

\frac{1}{4}

✅

-\frac{1}{4}

1

-1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiate: $2x-3y^{2}y'=y+xy'$ . Substituting $(1,1)$ gives $2-3y' = 1+y'$ . Solving, $1 = 4y'$ so $y' = \frac{1}{4}$ . The calculation shows that the tangent slope at $(1,1)$ is positive and equal to one‑quarter.

Q6. Compare the second derivative obtained by (i) implicit differentiation of $\sin y = x$ and (ii) differentiating $y=\arcsin x$ twice. Which statement is true?

A.Both methods give

-\dfrac{x}{(1-x^{2})^{3/2}}

✅

B.Implicit gives

-\dfrac{x}{(1-x^{2})^{3/2}}

while explicit gives

-\dfrac{x}{(1-x^{2})^{2}}

C.Implicit differentiation fails because

y

is not a global function

D.They differ only by a sign

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating $\sin y = x$ implicitly yields $y' = \frac{1}{\cos y}$ . Differentiating again gives $y'' = -\frac{\sin y}{\cos^{3} y}$ . Substituting $\sin y = x$ and $\cos y = \sqrt{1-x^{2}}$ results in $-\frac{x}{(1-x^{2})^{3/2}}$ . The same expression is obtained by directly differentiating $y=\arcsin x$ twice, confirming the equivalence.

Q7. The curve $x^{3}+y^{3}=3xy$ (Folium of Descartes) at $(1,1)$ satisfies $\frac{\partial F}{\partial y}=0$ . Which statement about differentiability is correct?

\frac{dy}{dx}=-1

\frac{dy}{dx}=0

\frac{dy}{dx}

is undefined because

\frac{\partial F}{\partial y}=0

✅

\frac{dy}{dx}=1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: At $(1,1)$ the partial derivative $\partial F/\partial y = 3y^{2}-3x = 0$ . The Implicit Function Theorem therefore does not guarantee a unique differentiable function $y(x)$ there, and the equation reduces to an identity $0=0$ . Consequently the slope cannot be determined by the usual formula, indicating that $\frac{dy}{dx}$ is undefined at that point.

Q8. If an implicit equation $F(x,y)=0$ has $\frac{\partial F}{\partial y}(x_{0},y_{0})=0$ but $\frac{\partial^{2}F}{\partial y^{2}}(x_{0},y_{0})\neq0$ , what can be concluded about a differentiable function $y(x)$ near $(x_{0},y_{0})$ ?

A.No differentiable function exists

B.A function exists but is not differentiable

C.A differentiable function may exist but the Implicit Function Theorem does not guarantee it ✅

D.A differentiable function definitely exists

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When the first‑order partial derivative with respect to $y$ vanishes, the standard Implicit Function Theorem cannot be applied. However, a non‑zero second‑order derivative suggests that the curve may still be locally solvable for $y$ as a function of $x$ ; the theorem simply does not provide a guarantee. Hence the existence of a differentiable function remains possible but unproven by the theorem.

Q9. For the relation $x^{2}y+3xy^{2}-x=3$ , find $\frac{dy}{dx}$ at $(x,y)=(1,1)$ .

-\frac{4}{7}

✅

\frac{4}{7}

-\frac{7}{4}

\frac{7}{4}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating gives $2xy + x^{2}y' + 3y^{2} + 6xyy' -1 =0$ . Substituting $(1,1)$ yields $4 + 7y' =0$ , so $y' = -\frac{4}{7}$ . This slope correctly reflects the tangent direction of the curve at the specified point.

Q10. Why must one verify that $\frac{\partial F}{\partial y}\neq0$ before applying implicit differentiation to $\cos(xy^{2}) = y$ ?

A.Because otherwise the derivative formula would involve division by zero ✅

B.Because

\frac{\partial F}{\partial y}

determines the curvature of the curve

C.Because the equation would have no solutions

D.Because it ensures

y

is a function of

x

locally

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When solving for $y'$ from $F_{x}+F_{y}y'=0$ , the term $F_{y}$ appears in the denominator. If $\frac{\partial F}{\partial y}=0$ at a point, the formula would require division by zero, making the computation invalid and indicating that the curve may fail to define $y$ as a function of $x$ there.

Q11. For the circle $x^{2}+y^{2}=100$ at the point $(6,8)$ , which equation represents the normal line?

y-8 = \frac{6}{8}(x-6)

y-8 = -\frac{6}{8}(x-6)

y-8 = \frac{8}{6}(x-6)

✅

y-8 = -\frac{8}{6}(x-6)

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The tangent slope is $-\frac{x}{y}=-\frac{6}{8}=-\frac{3}{4}$ ; the normal slope is the negative reciprocal, $\frac{4}{3}=\frac{8}{6}$ . Using point‑slope form with $(6,8)$ yields $y-8 = \frac{8}{6}(x-6)$ , which matches option C.

Q12. For $\sin(x^{2}y^{2}) = x$ , find $\frac{dy}{dx}$ at $(x,y) = (\pi/6,1)$ .

\displaystyle \frac{18}{\pi^{2}}\Bigl(\frac{1}{\cos(\pi^{2}/36)} - \frac{\pi}{3}\Bigr)

✅

\displaystyle \frac{18}{\pi^{2}}\Bigl(\frac{1}{\cos(\pi/6)} - \frac{\pi}{3}\Bigr)

\displaystyle \frac{9}{\pi^{2}}\Bigl(\frac{1}{\cos(\pi^{2}/36)} - \frac{\pi}{6}\Bigr)

\displaystyle \frac{18}{\pi^{2}}\Bigl(\frac{1}{\cos(\pi^{2}/36)} + \frac{\pi}{3}\Bigr)

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating gives $\cos(x^{2}y^{2})(2xy^{2}+2x^{2}y y')=1$ . Solving for $y'$ yields $y'=\frac{1/\cos(x^{2}y^{2})-2xy^{2}}{2x^{2}y}$ . Substituting $x=\pi/6$ and $y=1$ simplifies to the expression in option A.

Q13. For the curve defined by $x^{3}y^{3}=4$ at the point $(\sqrt[3]{2},\sqrt[3]{2})$ , which description of the local geometry is correct?

A.The curve has a vertical tangent

B.The curve has a horizontal tangent

C.The curve has a regular tangent with slope

-1

✅

D.The curve has a cusp

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Computing $\frac{dy}{dx}= -\frac{F_{x}}{F_{y}} = -\frac{3x^{2}y^{3}}{3x^{3}y^{2}} = -\frac{y}{x}$ . At the point where $x=y$ , the slope reduces to $-1$ . Both partial derivatives are non‑zero, so the Implicit Function Theorem applies and the curve possesses a smooth tangent line with slope $-1$ .

Q14. For the implicit relation $x^{2}y + y^{3}=6$ , determine $\frac{d^{2}y}{dx^{2}}$ at $(2,1)$ .

\displaystyle \frac{30}{343}

✅

-\displaystyle \frac{30}{343}

\displaystyle \frac{14}{9}

-\displaystyle \frac{14}{9}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: First derivative: $y' = -\frac{2xy}{x^{2}+3y^{2}}$ . At $(2,1)$ , $y'=-\frac{4}{7}$ . Using the formula $y'' = -\frac{F_{xx}+2F_{xy}y'+F_{yy}(y')^{2}}{F_{y}}$ with $F=x^{2}y+y^{3}-6$ gives \(

🔗 Related Topics (MCQs)

📝 Explicit vs implicit functions MCQs 📝 Implicit differentiation examples MCQs 📝 Derivatives of Logarithmic Functions MCQs 📝 Derivative of ln x and log base b MCQs 📝 Logarithmic differentiation technique MCQs

←Previous Topic:Implicit differentiation examples📝 Implicit differentiation differentiabilityDerivatives of Logarithmic Functions:Next Topic→