Differentiation of Implicit Functions MCQ Quiz - Objective Question with Answer for Differentiation of Implicit Functions - Download Free PDF

Calculation:

Given,

$(x+y{)}^{p+q}=x^p{y}^q$ and $p+q=10$.

Differentiate both sides with respect to $x$ implicitly:

$\frac{d}{dx}((x+y{)}^{p+q})=\frac{d}{dx}(x^p{y}^q)$

Left side:

$(p+q)(x+y{)}^{p+q-1}(1+\frac{dy}{dx})$

Right side (product rule):

$p{x}^{p-1}{y}^q+q{x}^p{y}^{q-1}\frac{dy}{dx}$

Rearrange to collect $\frac{dy}{dx}$ terms and use

$(x+y{)}^{p+q}=x^p{y}^q⟹(x+y{)}^{p+q-1}=\frac{x^p{y}^q}{x+y}$.

After cancellation of the common factor $py-qx$, you obtain:

∴ $\frac{dy}{dx}=\frac{y}{x}$

Hence, the correct answer is Option 1. 

Calculation:

Given,

$(x+y{)}^{p+q}=x^p{y}^q$

Differentiate implicitly w.r.t. $x$:

$(p+q)(x+y{)}^{p+q-1}(1+\frac{dy}{dx})=p{x}^{p-1}{y}^q+q{x}^p{y}^{q-1}\frac{dy}{dx}$

Rearrange to collect $\frac{dy}{dx}$:

$\frac{dy}{dx}[(p+q)(x+y{)}^{p+q-1}-q{x}^p{y}^{q-1}]=p{x}^{p-1}{y}^q-(p+q)(x+y{)}^{p+q-1}$

Use $(x+y{)}^{p+q-1}=\frac{x^p{y}^q}{x+y}$ to simplify:

$\frac{dy}{dx}=\frac{y}{x}$

∴ ${\displaystyle \frac{dy}{dx}=\frac{y}{x}}$, independent of $p$ and $q$.

Hence, the correct answer is Option 4.

Calculation:

Given,

$(x+y{)}^{p+q}=x^p{y}^q$ and $p+q=10$.

Differentiate both sides with respect to $x$ implicitly:

$\frac{d}{dx}((x+y{)}^{p+q})=\frac{d}{dx}(x^p{y}^q)$

Left side:

$(p+q)(x+y{)}^{p+q-1}(1+\frac{dy}{dx})$

Right side (product rule):

$p{x}^{p-1}{y}^q+q{x}^p{y}^{q-1}\frac{dy}{dx}$

Rearrange to collect $\frac{dy}{dx}$ terms and use

$(x+y{)}^{p+q}=x^p{y}^q⟹(x+y{)}^{p+q-1}=\frac{x^p{y}^q}{x+y}$.

After cancellation of the common factor $py-qx$, you obtain:

∴ $\frac{dy}{dx}=\frac{y}{x}$

Hence, the correct answer is Option 1. 

Calculation:

Given,

$(x+y{)}^{p+q}=x^p{y}^q$

Differentiate implicitly w.r.t. $x$:

$(p+q)(x+y{)}^{p+q-1}(1+\frac{dy}{dx})=p{x}^{p-1}{y}^q+q{x}^p{y}^{q-1}\frac{dy}{dx}$

Rearrange to collect $\frac{dy}{dx}$:

$\frac{dy}{dx}[(p+q)(x+y{)}^{p+q-1}-q{x}^p{y}^{q-1}]=p{x}^{p-1}{y}^q-(p+q)(x+y{)}^{p+q-1}$

Use $(x+y{)}^{p+q-1}=\frac{x^p{y}^q}{x+y}$ to simplify:

$\frac{dy}{dx}=\frac{y}{x}$

∴ ${\displaystyle \frac{dy}{dx}=\frac{y}{x}}$, independent of $p$ and $q$.

Hence, the correct answer is Option 4.

Concept:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{a}}^{\mathrm{x}}={\mathrm{a}}^{\mathrm{x}}\log \mathrm{a}$

Calculation:

Given 2x + 2y = 2x+y

We know that 2^a+b = 2^a⋅ 2^b

⇒ 2^x + 2^y = 2^x ⋅ 2^y

⇒ $\frac{2^{\mathrm{x}}+2^{\mathrm{y}}}{2^{\mathrm{x}}\cdot 2^{\mathrm{y}}}=1$

⇒ 2^-y + 2^-x = 1  ..(1)

Differentiating the above equation with respect to x:

⇒ (-2^{- y}$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ - 2^{- x}) log 2 = 0

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{2^{-\mathrm{x}}}{2^{-\mathrm{y}}}$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{1-2^{-\mathrm{y}}}{2^{-\mathrm{y}}}$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = - 2^y + 1

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 1 - 2^y

The required value of  $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ is 1 - 2^y .  

Calculation:

xe = e$({\mathrm{x}}^2+{\mathrm{y}}^2)$ 

Taking log on both sides, we get

⇒ log xe = log e${\mathrm{x}}^2+{\mathrm{y}}^2$ 

⇒ e log x = (x² + y²) log e             

[∵ log mⁿ = n log m]

⇒ e log x = x2 + y2                       

[∵ log e = 1]

Differentiating w.r.t x, we get

⇒ e($\frac{1}{\mathrm{x}}$) = $2\mathrm{x}+2\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

⇒ $\frac{\mathrm{e}}{\mathrm{x}}-2\mathrm{x}=2\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

∴ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{e}-2{\mathrm{x}}^2}{2\mathrm{x}\mathrm{y}}$

Concept:

Calculus:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left({\mathrm{a}}^{\mathrm{x}}\right)={\mathrm{a}}^{\mathrm{x}}(\log \mathrm{a})$.
•  

Chain Rule of Derivatives:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{d}\mathrm{g}(\mathrm{x})}\mathrm{f}(\mathrm{g}(\mathrm{x}))\times \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{g}(\mathrm{x})$.

Calculation:

Given expression is:

3x + 3y = 3x + y.

Differentiating w.r.t. x and using the chain rule of derivatives, we get:

⇒ 3x (log 3) + 3y (log 3) $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 3x + y (log 3) (1 + $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$)

⇒ 3x + 3y $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 3x + y + 3x + y $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{3^{\mathrm{x}+\mathrm{y}}-3^{\mathrm{x}}}{3^{\mathrm{y}}-3^{\mathrm{x}+\mathrm{y}}}$

Calculation:

Here, 

$2{\mathrm{x}}^3-3{\mathrm{y}}^2=7$

Differentiating w.r.t. x, we get

$$\begin{gathered} 6{\mathrm{x}}^2-6\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=0 \\ \Rightarrow {\mathrm{x}}^2-\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=0 \\ \Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{{\mathrm{x}}^2}{\mathrm{y}} \end{gathered}$$

Hence, option (3) is correct.

Calculation:

y + sin-1 (1 - x2) = ex 

y = e^x - sin^-1 (1 - x²)

Differentiating w.r.t x, we get

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}-\frac{1}{\sqrt{1-(1-{\mathrm{x}}^2{)}^2}}(-2\mathrm{x})$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}+\frac{2\mathrm{x}}{\sqrt{1-(1-2{\mathrm{x}}^2+{\mathrm{x}}^4)}}$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}+\frac{2\mathrm{x}}{\sqrt{2{\mathrm{x}}^2-{\mathrm{x}}^4}}$

$\text{boldsymbol}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}+\frac{2}{\sqrt{2-{\mathrm{x}}^2}}$

Concept:

$\frac{\mathrm{d}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{d}\mathrm{x}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$

Calculation:

Given: $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

Differentiating with respect to x, we get

$\Rightarrow \frac{2\mathrm{x}}{{\mathrm{a}}^2}+\frac{2\mathrm{y}}{{\mathrm{b}}^2}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=0$

$\Rightarrow \frac{2\mathrm{y}}{{\mathrm{b}}^2}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{2\mathrm{x}}{{\mathrm{a}}^2}$

$\therefore \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{{\mathrm{b}}^2\mathrm{x}}{{\mathrm{a}}^2\mathrm{y}}$

Concept:

$\frac{\mathrm{d}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{d}\mathrm{x}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$

Calculation:

Given: y2 = 4ax

Differentiating with respect to x, we get

$\Rightarrow 2\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=4\mathrm{a}$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{4\mathrm{a}}{2\mathrm{y}}$

$\therefore \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{2\mathrm{a}}{\mathrm{y}}$

Given

y = 3e^2x + 2e^3x

Formula used

d(xⁿ)/dx = nx^n-1

Solution

⇒dy/dx = 3e^2x(2) + 2e^3x(3)

⇒dy/dx = 6(e^2x + e^3x)

⇒d²y/dx² = 6(2e^2x+3e^3x)

As asked in the question,

⇒  $\frac{d^{2}y}{d{x}^2}$ - 5 $\frac{dy}{dx}$ + 6y =

⇒ 12e^2x + 18e^3x − 30e^2x − 30e^3x + 18e^2x + 12e^3x

⇒ 0.

The correct option is 4.

Concept:

Chain Rule (Differentiation by substitution): If y is a function of u and u is a function of x

• $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{u}}\times \frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}$

Calculation:

Given y2 + x2 + 3x + 5 = 0

Differentiating with respect to x, we get

2y $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ + 2x +3(1) + 0 = 0

2y$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ + 2x + 3 = 0 

2y $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = -(2x + 3)

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{2\mathrm{x}+3}{2\mathrm{y}}$

Now at (0, -3)

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{2(0)+3}{2(-3)}$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{3}{(-6)}$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{2}$ = 0.5

Calculation:

Given function is y² = 2x;

Differentiating with respect to x on both sides,

2yy' = 2x ⇒ yy’ = 1   - (1)

⇒ y’ = 1/y;

Differentiating (1) again with respect to x on both sides,

⇒ y . y” + y’. y’ = 0   - (2)

$\Rightarrow y^{″}=-\frac{{\left(y^{\prime }\right)}^2}{y}=-\frac{1}{y^3}$

Concept:

cos2x = 2cos²x - 1

sin2x = 2sin x cos x

Calculation:

Here, y = cos2 x2

Let, x² = t 

Differentiating with respect to x, we get

⇒2xdx = dt 

⇒ dt/dx = 2x ....(1)

y = cos²t

=$\frac{\cos 2\mathrm{t}+1}{2}=\frac{\cos 2\mathrm{t}}{2}+\frac{1}{2}$

$$\begin{gathered} \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}(\cos 2\mathrm{t})\frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{x}}+0 \\ =\frac{1}{2}(-2\sin 2\mathrm{t})\frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{x}}\cdots (\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}(1)) \end{gathered}$$

= - sin2x² × 2x

= -4x cos x2 sin x2

Hence, option (2) is correct.