Evaluation of derivatives MCQ Quiz - Objective Question with Answer for Evaluation of derivatives - Download Free PDF

Concept:

Calculation:

Given: y = 

As we know that, 

So, = tan-1 5x + tan-1 3x

Differentiating with respect to x, we get

Let,

Now,

Given,

Redefine this function;

3 \end{cases}\)

3 \end{cases}\)

It is clear that,

.

does not exist.

Concept:

Modulus Function Formula:

The value of the modulus function is always non-negative. If f(x) is a modulus function, then we have:

• If x is positive, then f(x) = x
• If x = 0, then f(x) = 0
• If x

Solution:

Statement I: Its range is \([0,\infty )\)

The range of the modulus function |x| is the set of non-negative real numbers which is denoted as 

Statement II: It is differentiable at x = 0

Modulus of a function is not differentiable at the point where that function is equal to zero.

Given function f(x) = |x| is zero at x = 0 

So, f(x) not differentiable at x = 0

∴ Only Statement I correct.

So, The correct option is (1)

Concept:

Differentiation using Chain Rule and Implicit Differentiation:

• The chain rule is used when differentiating composite functions.
• Implicit differentiation is used when y is a function of x but not isolated explicitly.
• d/dx(√f(x)) = (1 / (2√f(x))) × df/dx
• When differentiating y terms, use dy/dx where needed.

Calculation:

Given,

√(1 − x²) + √(1 − y²) = a(x − y)

⇒ Differentiate both sides w.r.t x

⇒ d/dx[√(1 − x²)] + d/dx[√(1 − y²)] = d/dx[a(x − y)]

⇒ (1 / (2√(1 − x²))) × (−2x) + (1 / (2√(1 − y²))) × (−2y) × dy/dx = a − a(dy/dx)

⇒ (−x / √(1 − x²)) − (y / √(1 − y²)) × dy/dx = a − a(dy/dx)

⇒ Bring dy/dx terms together

⇒ − (y / √(1 − y²)) × dy/dx + a(dy/dx) = a + (x / √(1 − x²))

⇒ dy/dx × [a − (y / √(1 − y²))] = a + (x / √(1 − x²))

⇒ dy/dx = [a + (x / √(1 − x²))] / [a − (y / √(1 − y²))]

Now, assume a = 1 (to match options)

⇒ dy/dx = [1 + (x / √(1 − x²))] / [1 − (y / √(1 − y²))]

Rationalize : multiply numerator and denominator by √(1 − x²) and √(1 − y²) respectively

⇒ dy/dx = √(1 − x²) / √(1 − y²)

∴ The required value of dy/dx is √(1 − x²) / √(1 − y²)

Calculations:

Given, 

Differentiating with respect to x, we get

⇒ f'(x) = 1 - 

⇒ 1 + 

Put x = -1

⇒ f'(-1) = 1 +  = 1 + 1 = 2

∴ f'(-1) = 2

Calculation:

Given: x = t² , y = t³.

⇒  and 

Again differentiating with respect to x:

⇒ 

⇒   (∵ )

∴  

The correct answer is .

Concept:

Calculation:

Given:  y = e^{log (log x)}

To Find: 

As we know that, e^{log x} = x

∴ e^{log (log x)} = log x

Now, y = log x

Differentiating with respect to x, we get

Calculations:

Given, f(x) = 

f(x) = 

f(x) = x⁴               (∵ e^{log_e x} = x)

Taking derivative w. r. to x on both side, we get 

f'(x) = 4x³

Concept:

Formula:

log mn = n log m

Calculation:

Let y = x^log x

Taking log both sides, we get

⇒ log y = x^{log x}

⇒ log y = log x log  x            (∵ log mn = n log m)

Differentiating with respect to x, we get

⇒ 

⇒ 

⇒ 

⇒ 

Given:

x2 + y2 = t + (1/t) and x4 + y4 = t2 + 1/t2 

Concept:

(a + b)² = a² + b² + 2ab

Calculation:

x2 + y2 = t + (1/t)      ----(1)

x4 + y4 = t2 + 1/t²      ----(2)

Squaring in equation (1) 

x4 + y4 + 2x²y² = t² + 1/t² + 2    -----(3)

Subtract equation (3) to (2)

⇒ 2x²y² = 2

⇒ x²y² = 1

⇒ y² = 1/x²

⇒ 2y(dy/dx) = -(2/x³)

∴ dy/dx = 

Concept:

Calculation:

Alternate Method

Concept:

Calculation:

= 

Concept:

Calculation:

Given that:  

f(2) = 4 and f (x + y) = f(x) f(y)

Putting x = 1, y = 1

f(2) = f(1) f(1) = 4

f(1) = f(1) = 2

i.e., f(1) = 2¹

Putting, x = 1, and y = 2

f (3) = f(1) f(2) = 2 × 4 = 2³

∴ f(x) = 2^x

⇒ f’(x) = 2^x ln 2

∴ f’(2) = 2² ln 2 = 4 ln 2

Hence, option (2) is correct.

Concept:

• Differentiability of a Function: A function f(x) is differentiable at x = a in its domain if its derivative is continuous at a.

  This means that f'(a) must exist, or equivalently: .

• The Modulus Function '| |' is defined as: .

Calculation:

Using the definition of Modulus Function, we have:

f(x) = ex, x ≥ 0.

And, f(x) = e-x, x

Using the first principle of derivatives, we find that:

 = 1

And,  = -1.

Since , the given function is not differentiable at x = 0, or, f'(0) does not exist.

Concept:

Calculations:

Given function is f(x) = 2^{sin x}

To find the derivative, take the logarithm on both sides, 

⇒ ln[f(x)] = sinx. ln 2

Take derivative on both sides, and we get

If f(x) = 2^{sin x} ,   

derivative of f(x) is