Differentiation of Implicit Functions MCQ Quiz in मल्याळम - Objective Question with Answer for Differentiation of Implicit Functions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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:

Given:

$2x^2-3xy+4y^2+2x-3y+4=0$

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

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

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

⇒ $(8y-3x-3)\frac{dy}{dx}=3y-4x-2$

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

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

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

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

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

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

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

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

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

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

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