Question 1

Passage: Let (e^y )^x-y=0, where y is a function of x whose domain is (0,10]. Question: What is dy dx equal to, given that y=y_0 when x=1 ?

Accepted Answer

Given the equation (e^y)^x - y = 0, we can rewrite it as: e^ xy = y Taking the natural logarithm on both sides: xy = y Differentiating both sides with respect to x using the product rule and chain rule: y + x dy dx = 1 y dy dx Rearranging the terms to solve for dy dx : y = ( 1 y - x ) dy dx y = ( 1 - xy y ) dy dx dy dx = y^2 1 - xy We are given that at x = 1, y = y_0. Substituting these values into the derivative: dy dx = y_0^2 1 - y_0 From the original equation e^ xy = y, substituting x = 1 and

Question 2

If [4] y =x+ x^2+4 , then what is x^2+4 \, dy dx equal to ?

Accepted Answer

Given y^ 1/4 = x + x^2+4 Differentiating both sides with respect to x: 1 4 y^ -3/4 dy dx = 1 + 1 2 x^2+4 (2x) 1 4 y^ -3/4 dy dx = 1 + x x^2+4 1 4 y^ -3/4 dy dx = x^2+4 + x x^2+4 Substituting x + x^2+4 = y^ 1/4 : 1 4 y^ -3/4 dy dx = y^ 1/4 x^2+4 x^2+4 dy dx = 4 y^ 1/4 y^ 3/4 x^2+4 dy dx = 4y Answer: 4y

Question 3

Passage: Let (e^y )^x-y=0, where y is a function of x whose domain is (0,10]. Question: What is dy dx equal to ?

Accepted Answer

Given equation is (e^y)^x - y = 0 e^ xy = y Taking natural logarithm on both sides: xy = (y) Differentiating both sides with respect to x: y + x dy dx = 1 y dy dx y = ( 1 y - x ) dy dx y = ( 1 - xy y ) dy dx dy dx = y^2 1 - xy

Question 4

For the following two (02) items: Consider the function f(x) = cases 4(5^x), & x What is f'(-1) equal to?

Accepted Answer

For x Differentiating with respect to x, we get: f'(x) = 4(5^x 5) To find f'(-1), substitute x = -1 into the derivative: f'(-1) = 4(5^ -1 5) f'(-1) = 4 5 5 Answer: 4 5 5

Question 5

Consider the following for the two (02) items that follow: Let x = - and y = ^4 - ^4 . What is ( dy dx )^2 equal to?

Accepted Answer

Given x = - and y = ^4 - ^4 . Differentiating x with respect to : dx d = + dx d = ( + ) Differentiating y with respect to : dy d = 4 ^3 ( ) - 4 ^3 (- ) dy d = 4 ^4 + 4 ^3 dy d = 4 ( ^4 + ^4 ) Now, dy dx = dy d dx d = 4 ( ^4 + ^4 ) ( + ) = 4( ^4 + ^4 ) + Squaring both sides: ( dy dx )^2 = 16( ^4 + ^4 )^2 ( + )^2 Using the algebraic identity (a+b)^2 = (a-b)^2 + 4ab, we can rewrite the numerator and denominator: ( + )^2 = ( - )^2 + 4 = x^2 + 4 ( ^4 + ^4 )^2 = ( ^4 - ^4 )^2 + 4 ^4 ^4 = y^2 + 4 Subst

Question 6

Consider the following for the two (02) items that follow: Let (x+y)^ p+q = x^p y^q, where p, q are positive integers. The derivative of y with respect to x

Accepted Answer

Given (x+y)^ p+q = x^p y^q Taking the natural logarithm on both sides: (p+q) (x+y) = p x + q y Differentiating both sides with respect to x: p+q x+y (1 + dy dx ) = p x + q y dy dx Rearranging the terms to solve for dy dx : p+q x+y - p x = ( q y - p+q x+y ) dy dx Simplifying the left side: x(p+q) - p(x+y) x(x+y) = px + qx - px - py x(x+y) = qx - py x(x+y) Simplifying the right side: q(x+y) - y(p+q) y(x+y) = qx + qy - py - qy y(x+y) = qx - py y(x+y) Equating the two sides: qx - py x(x+y) = qx - py

Question 7

Consider the following for the two (02) items that follow: Let y = ^ -1 (x - 4x^3 27 ). What is dy dx equal to?

Accepted Answer

Let x = 3 . Then = ^ -1 ( x 3 ). Substituting x = 3 into the given expression: y = ^ -1 (3 - 4(3 )^3 27 ) y = ^ -1 (3 - 4(27 ^3 ) 27 ) y = ^ -1 (3 - 4 ^3 ) Using the trigonometric identity (3 ) = 3 - 4 ^3 : y = ^ -1 ( 3 ) Assuming principal values, we get: y = 3 y = 3 ^ -1 ( x 3 ) Differentiating with respect to x using the chain rule: dy dx = 3 1 1 - ( x 3 )^2 d dx ( x 3 ) dy dx = 3 1 1 - x^2 9 1 3 dy dx = 1 9 - x^2 9 dy dx = 3 9 - x^2 Answer: 3 9-x^2

Question 8

For the following three (03) items: Let y = f(x) = x ^ -1 x 1 - x^2 + 1 - x^2 . If x = , then what is dy dx equal to?

Accepted Answer

Given y = x ^ -1 x 1 - x^2 + 1 - x^2 Substitute x = into the expression for y: y = 1 - ^2 + 1 - ^2 y = + ( ) y = + ( ) Differentiating y with respect to : dy d = d d ( ) + d d ( ( )) dy d = ( + ^2 ) + 1 (- ) dy d = + ^2 - = ^2 Differentiating x = with respect to : dx d = Using the chain rule to find dy dx : dy dx = dy d dx d = ^2 dy dx = ^3

Question 9

Consider the following for the two (02) items that follow: Let (x+y)^ p+q = x^p y^q, where p, q are positive integers. If p + q = 10, then what is dy dx equal to?

Accepted Answer

Given (x+y)^ p+q = x^p y^q Taking natural logarithm on both sides: (p+q) (x+y) = p x + q y Differentiating both sides with respect to x: p+q x+y (1 + dy dx ) = p x + q y dy dx Rearranging the terms to solve for dy dx : p+q x+y - p x = ( q y - p+q x+y ) dy dx x(p+q) - p(x+y) x(x+y) = ( q(x+y) - y(p+q) y(x+y) ) dy dx qx - py x(x+y) = ( qx - py y(x+y) ) dy dx dy dx = y x The result is independent of the value of p+q. Answer: y x

Question 10

For the following three (03) items: Let y = f(x) = x ^ -1 x 1 - x^2 + 1 - x^2 . What is d^2y dx^2 at x = 0 equal to?

Accepted Answer

Given y = x ^ -1 x 1 - x^2 + 1 - x^2 We can rewrite the function as: y = x ^ -1 x 1 - x^2 + 1 2 (1 - x^2) Differentiating with respect to x using the quotient and chain rules: dy dx = ( ^ -1 x + x 1 - x^2 ) 1 - x^2 - x ^ -1 x ( -x 1 - x^2 ) 1 - x^2 + 1 2 ( -2x 1 - x^2 ) dy dx = 1 - x^2 ^ -1 x + x + x^2 ^ -1 x 1 - x^2 1 - x^2 - x 1 - x^2 dy dx = (1 - x^2) ^ -1 x + x 1 - x^2 + x^2 ^ -1 x (1 - x^2)^ 3/2 - x 1 - x^2 dy dx = ^ -1 x + x 1 - x^2 (1 - x^2)^ 3/2 - x 1 - x^2 dy dx = ^ -1 x (1 - x^2)^ 3/2

Question 11

Consider the following for the two (02) items that follow: Let x = - and y = ^4 - ^4 . What is ( x^2+4 y^2+4 ) dy dx [(x^2+4) d^2y dx^2 - 16y ] equal to?

Accepted Answer

Given x = - and y = ^4 - ^4 . First, we simplify x^2 + 4 and y^2 + 4: x^2 + 4 = ( - )^2 + 4 = ^2 + ^2 - 2 + 4 = ( + )^2 y^2 + 4 = ( ^4 - ^4 )^2 + 4 = ^8 + ^8 - 2 + 4 = ( ^4 + ^4 )^2 Differentiating x and y with respect to : dx d = + = ( + ) dy d = 4 ^3 ( ) - 4 ^3 (- ) = 4 ( ^4 + ^4 ) Now, finding dy dx : dy dx = dy d dx d = 4 ( ^4 + ^4 ) ( + ) = 4( ^4 + ^4 ) + Squaring both sides: ( dy dx )^2 = 16( ^4 + ^4 )^2 ( + )^2 = 16(y^2+4) x^2+4 Rearranging the equation: (x^2+4) ( dy dx )^2 = 16(y^2+4) Di

Differentiation — Previous Year Questions & Solutions