Differentiation of Implicit Functions MCQ Quiz in मराठी - Objective Question with Answer for Differentiation of Implicit Functions - मोफत PDF डाउनलोड करा

Calculation:

Given,

\((x+y)^{p+q} = x^p\,y^q\)

Differentiate implicitly w.r.t. \(x\):

\((p+q)(x+y)^{p+q-1}\bigl(1+\tfrac{dy}{dx}\bigr) = p\,x^{p-1}y^q \;+\; q\,x^p\,y^{q-1}\tfrac{dy}{dx}\)

Rearrange to collect \(\tfrac{dy}{dx}\):

\(\tfrac{dy}{dx}\bigl[(p+q)(x+y)^{p+q-1} - q\,x^p\,y^{q-1}\bigr] = p\,x^{p-1}y^q - (p+q)(x+y)^{p+q-1}\)

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

\(\tfrac{dy}{dx} = \tfrac{y}{x}\)

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

Hence, the correct answer is Option 4.

Concept:

• \(\rm \frac{d}{dx}a^x=a^x\log a\)

Calculation:

Given 2x + 2y = 2x+y

We know that 2^a+b = 2^a⋅ 2^b

⇒ 2^x + 2^y = 2^x ⋅ 2^y

⇒ \(\rm \frac{2^x+2^y}{2^x\cdot2^y}=1\)

⇒ 2^-y + 2^-x = 1  ..(1)

Differentiating the above equation with respect to x:

⇒ (-2^{- y}\(\rm\frac{dy}{dx}\) - 2^{- x}) log 2 = 0

⇒ \(\rm \frac{dy}{dx}=-\frac{2^{-x}}{2^{-y}}\)

⇒ \(\rm \frac{dy}{dx}=-\frac{1-2^{-y}}{2^{-y}}\)

⇒ \(\rm \frac{dy}{dx}\) = - 2^y + 1

⇒ \(\rm \frac{dy}{dx}\) = 1 - 2^y

The required value of  \(\rm \frac{dy}{dx}\) is 1 - 2^y .  

Calculation:

Given:

\(2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0\)

Differentiating the given equation with respect to x, we get:

⇒ \(4x - 3\left(x\frac{dy}{dx} + y\right) + 8y\frac{dy}{dx} + 2 - 3\frac{dy}{dx} = 0\)

⇒ \(4x - 3x\frac{dy}{dx} - 3y + 8y\frac{dy}{dx} + 2 - 3\frac{dy}{dx} = 0\)

⇒ \((8y - 3x - 3)\frac{dy}{dx} = 3y - 4x - 2\)

⇒ \(\frac{dy}{dx} = \frac{3y - 4x - 2}{8y - 3x - 3}\) 

⇒ \(\left(\frac{dy}{dx}\right)_{(3,2)} = \frac{3(2) - 4(3) - 2}{8(2) - 3(3) - 3}\)

⇒ \(\left(\frac{dy}{dx}\right)_{(3,2)} = \frac{6 - 12 - 2}{16 - 9 - 3}\)

⇒ \(\left(\frac{dy}{dx}\right)_{(3,2)} = \frac{-8}{4}\)

⇒ \(\left(\frac{dy}{dx}\right)_{(3,2)} = -2\)

∴ \(\left(\frac{dy}{dx}\right)_{(3,2)} = -2\)

Hence option 3 is correct

Calculation

\(y = \alpha x^2 + \cos y + \beta\)

Differentiate both sides with respect to x:

⇒ \(\frac{dy}{dx} = 2\alpha x - \sin y \frac{dy}{dx}\)

⇒ \(\frac{dy}{dx} (1 + \sin y) = 2\alpha x\)

⇒ \(\frac{dy}{dx} = \frac{2\alpha x}{1 + \sin y}\)

At (1, 0):

⇒ \(2 = \frac{2\alpha (1)}{1 + \sin 0}\)

⇒ \(2 = \frac{2\alpha}{1 + 0}\)

⇒ \(2 = 2\alpha\)

⇒ \(\alpha = 1\)

Substitute (1, 0) in the given equation:

⇒ \(0 = \alpha (1)^2 + \cos 0 + \beta\)

⇒ \(0 = \alpha + 1 + \beta\)

⇒ \(0 = 1 + 1 + \beta\)

⇒ \(\beta = -2\)

\(\alpha \beta = 1 \times (-2)\)

⇒ \(\alpha \beta = -2\)

\(\therefore \alpha \beta = -2\)

Hence option 4 is correct

\( \cfrac { d(\log(\sec\theta +\tan\theta )) }{ d(\sec\theta ) } =\cfrac { (\cfrac { \sec\theta \tan\theta +{ \sec }^{ 2 }\theta }{ \sec\theta +\tan\theta } ) }{ \sec\theta \tan\theta } =\cfrac { 1 }{ \tan\theta } =\cot\theta =\cot(\cfrac { \pi }{ 4 })=1 \)

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

• \(\rm {dy\over dx} = {dy\over du}× {du\over dx}\)

Calculation:

Given y2 + x2 + 3x + 5 = 0

Differentiating with respect to x, we get

2y \(\rm dy\over dx\) + 2x +3(1) + 0 = 0

2y\(\rm dy\over dx\) + 2x + 3 = 0 

2y \(\rm dy\over dx\) = -(2x + 3)

\(\rm {dy\over dx} = -{2x+3\over2y}\)

Now at (0, -3)

\(\rm {dy\over dx} = -{2(0)+3\over2(-3)}\)

\(\rm {dy\over dx} = -{3\over(-6)}\)

\(\rm {dy\over dx} = {1\over2}\) = 0.5