Answer :
Let's solve each of the questions one by one.
Question 16:
Given [tex]\( y = \sin(\ln(x^5)) \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex].
Step 1: Differentiate the inside function [tex]\( \ln(x^5) \)[/tex].
[tex]\[ \text{Let} \ u = \ln(x^5) = 5 \ln(x). \][/tex]
[tex]\[ \frac{du}{dx} = 5 \cdot \frac{1}{x} = \frac{5}{x} \][/tex]
Step 2: Differentiate the outer function [tex]\( \sin(u) \)[/tex].
[tex]\[ y = \sin(u) \][/tex]
[tex]\[ \frac{dy}{du} = \cos(u) \][/tex]
Step 3: Apply the chain rule:
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(\ln(x^5)) \cdot \frac{5}{x} = \frac{5}{x} \cos(\ln(x^5)) \][/tex]
So, the correct answer is:
(C) [tex]\(\frac{5}{x} \cos(\ln(x^5))\)[/tex].
Question 17:
Given [tex]\( y = \frac{(1-x)^{20}}{(1+x)^{25}} \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex].
[tex]\[ y = (1-x)^{20} (1+x)^{-25} \][/tex]
Step 1: Use the product rule and chain rule. Let [tex]\( u = (1-x)^{20} \)[/tex] and [tex]\( v = (1+x)^{-25} \)[/tex], so [tex]\( y = u \cdot v \)[/tex].
First, compute [tex]\( \frac{du}{dx} \)[/tex]:
[tex]\[ u = (1-x)^{20} \][/tex]
[tex]\[ \frac{du}{dx} = 20(1-x)^{19} \cdot (-1) = -20(1-x)^{19} \][/tex]
Next, compute [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ v = (1+x)^{-25} \][/tex]
[tex]\[ \frac{dv}{dx} = -25(1+x)^{-26} \cdot 1 = -25(1+x)^{-26} \][/tex]
Step 2: Apply the product rule:
[tex]\[ \frac{dy}{dx} = v \frac{du}{dx} + u \frac{dv}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = (1+x)^{-25} \cdot (-20(1-x)^{19}) + (1-x)^{20} \cdot (-25(1+x)^{-26}) \][/tex]
[tex]\[ \frac{dy}{dx} = -20(1-x)^{19} (1+x)^{-25} - 25(1-x)^{20} \cdot (1+x)^{-26} \][/tex]
[tex]\[ \frac{dy}{dx} = y \left( -\frac{20}{1-x} - \frac{25}{1+x} \right) \][/tex]
So the correct answer is:
(D) [tex]\( -y \left(\frac{20}{1-x} + \frac{25}{1+x}\right) \)[/tex].
Question 18:
Given [tex]\( \cos(y^3) = x^3 \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex].
Step 1: Differentiate both sides implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (\cos(y^3)) = \frac{d}{dx} (x^3) \][/tex]
Step 2: Apply the chain rule on the left side and direct differentiation on the right:
[tex]\[ -\sin(y^3) \cdot \frac{d}{dx} (y^3) = 3x^2 \][/tex]
[tex]\[ -\sin(y^3) \cdot 3y^2 \cdot \frac{dy}{dx} = 3x^2 \][/tex]
Step 3: Solve for [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{3x^2}{-3y^2 \sin(y^3)} = \frac{-x^2}{y^2 \sin(y^3)} \][/tex]
So the correct answer is:
(A) [tex]\(\frac{-x^2}{y^2 \sin(y^3)}\)[/tex].
Question 19:
Given [tex]\( x^3 y^3 + x = y \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex] at [tex]\( (1,1) \)[/tex].
Step 1: Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (x^3 y^3 + x) = \frac{d}{dx} y \][/tex]
Step 2: Apply the product rule on the left side and direct differentiation on the right:
[tex]\[ 3x^2 y^3 + x^3 \cdot 3y^2 \cdot \frac{dy}{dx} + 1 = \frac{dy}{dx} \][/tex]
Step 3: Collect like terms and solve for [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 3x^2 y^3 + 1 = \frac{dy}{dx} (1 - 3x^3 y^2) \][/tex]
Step 4: Evaluate at [tex]\( (1,1) \)[/tex]:
[tex]\[ 3(1)^2 (1)^3 + 1 = \frac{dy}{dx} (1 - 3(1)^3 (1)^2) \][/tex]
[tex]\[ 3 + 1 = \frac{dy}{dx} (1 - 3) \][/tex]
[tex]\[ 4 = -2 \frac{dy}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{4}{-2} = -2 \][/tex]
So the correct answer is:
(A) [tex]\(-2\)[/tex].
Question 20:
Given [tex]\( y = 4 \sin(2t) \)[/tex] and [tex]\( x = 8 e^t \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex] at [tex]\( t=0 \)[/tex].
Step 1: Differentiate [tex]\( y \)[/tex] and [tex]\( x \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ \frac{dy}{dt} = 4 \cdot 2 \cos(2t) = 8 \cos(2t) \][/tex]
[tex]\[ \frac{dx}{dt} = 8 e^t \][/tex]
Step 2: Find [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{8 \cos(2t)}{8 e^t} = \frac{\cos(2t)}{e^t} \][/tex]
Step 3: Evaluate at [tex]\( t=0 \)[/tex]:
[tex]\[ \frac{dy}{dx} \bigg|_{t=0} = \frac{\cos(2 \cdot 0)}{e^0} = \frac{\cos(0)}{1} = 1 \][/tex]
So the correct answer is:
(B) [tex]\( 1 \)[/tex].
Question 16:
Given [tex]\( y = \sin(\ln(x^5)) \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex].
Step 1: Differentiate the inside function [tex]\( \ln(x^5) \)[/tex].
[tex]\[ \text{Let} \ u = \ln(x^5) = 5 \ln(x). \][/tex]
[tex]\[ \frac{du}{dx} = 5 \cdot \frac{1}{x} = \frac{5}{x} \][/tex]
Step 2: Differentiate the outer function [tex]\( \sin(u) \)[/tex].
[tex]\[ y = \sin(u) \][/tex]
[tex]\[ \frac{dy}{du} = \cos(u) \][/tex]
Step 3: Apply the chain rule:
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(\ln(x^5)) \cdot \frac{5}{x} = \frac{5}{x} \cos(\ln(x^5)) \][/tex]
So, the correct answer is:
(C) [tex]\(\frac{5}{x} \cos(\ln(x^5))\)[/tex].
Question 17:
Given [tex]\( y = \frac{(1-x)^{20}}{(1+x)^{25}} \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex].
[tex]\[ y = (1-x)^{20} (1+x)^{-25} \][/tex]
Step 1: Use the product rule and chain rule. Let [tex]\( u = (1-x)^{20} \)[/tex] and [tex]\( v = (1+x)^{-25} \)[/tex], so [tex]\( y = u \cdot v \)[/tex].
First, compute [tex]\( \frac{du}{dx} \)[/tex]:
[tex]\[ u = (1-x)^{20} \][/tex]
[tex]\[ \frac{du}{dx} = 20(1-x)^{19} \cdot (-1) = -20(1-x)^{19} \][/tex]
Next, compute [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ v = (1+x)^{-25} \][/tex]
[tex]\[ \frac{dv}{dx} = -25(1+x)^{-26} \cdot 1 = -25(1+x)^{-26} \][/tex]
Step 2: Apply the product rule:
[tex]\[ \frac{dy}{dx} = v \frac{du}{dx} + u \frac{dv}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = (1+x)^{-25} \cdot (-20(1-x)^{19}) + (1-x)^{20} \cdot (-25(1+x)^{-26}) \][/tex]
[tex]\[ \frac{dy}{dx} = -20(1-x)^{19} (1+x)^{-25} - 25(1-x)^{20} \cdot (1+x)^{-26} \][/tex]
[tex]\[ \frac{dy}{dx} = y \left( -\frac{20}{1-x} - \frac{25}{1+x} \right) \][/tex]
So the correct answer is:
(D) [tex]\( -y \left(\frac{20}{1-x} + \frac{25}{1+x}\right) \)[/tex].
Question 18:
Given [tex]\( \cos(y^3) = x^3 \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex].
Step 1: Differentiate both sides implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (\cos(y^3)) = \frac{d}{dx} (x^3) \][/tex]
Step 2: Apply the chain rule on the left side and direct differentiation on the right:
[tex]\[ -\sin(y^3) \cdot \frac{d}{dx} (y^3) = 3x^2 \][/tex]
[tex]\[ -\sin(y^3) \cdot 3y^2 \cdot \frac{dy}{dx} = 3x^2 \][/tex]
Step 3: Solve for [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{3x^2}{-3y^2 \sin(y^3)} = \frac{-x^2}{y^2 \sin(y^3)} \][/tex]
So the correct answer is:
(A) [tex]\(\frac{-x^2}{y^2 \sin(y^3)}\)[/tex].
Question 19:
Given [tex]\( x^3 y^3 + x = y \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex] at [tex]\( (1,1) \)[/tex].
Step 1: Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (x^3 y^3 + x) = \frac{d}{dx} y \][/tex]
Step 2: Apply the product rule on the left side and direct differentiation on the right:
[tex]\[ 3x^2 y^3 + x^3 \cdot 3y^2 \cdot \frac{dy}{dx} + 1 = \frac{dy}{dx} \][/tex]
Step 3: Collect like terms and solve for [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 3x^2 y^3 + 1 = \frac{dy}{dx} (1 - 3x^3 y^2) \][/tex]
Step 4: Evaluate at [tex]\( (1,1) \)[/tex]:
[tex]\[ 3(1)^2 (1)^3 + 1 = \frac{dy}{dx} (1 - 3(1)^3 (1)^2) \][/tex]
[tex]\[ 3 + 1 = \frac{dy}{dx} (1 - 3) \][/tex]
[tex]\[ 4 = -2 \frac{dy}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{4}{-2} = -2 \][/tex]
So the correct answer is:
(A) [tex]\(-2\)[/tex].
Question 20:
Given [tex]\( y = 4 \sin(2t) \)[/tex] and [tex]\( x = 8 e^t \)[/tex], we want to find [tex]\( \frac{dy}{dx} \)[/tex] at [tex]\( t=0 \)[/tex].
Step 1: Differentiate [tex]\( y \)[/tex] and [tex]\( x \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ \frac{dy}{dt} = 4 \cdot 2 \cos(2t) = 8 \cos(2t) \][/tex]
[tex]\[ \frac{dx}{dt} = 8 e^t \][/tex]
Step 2: Find [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{8 \cos(2t)}{8 e^t} = \frac{\cos(2t)}{e^t} \][/tex]
Step 3: Evaluate at [tex]\( t=0 \)[/tex]:
[tex]\[ \frac{dy}{dx} \bigg|_{t=0} = \frac{\cos(2 \cdot 0)}{e^0} = \frac{\cos(0)}{1} = 1 \][/tex]
So the correct answer is:
(B) [tex]\( 1 \)[/tex].
