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QUESTION: 1

The dervative of y

Solution:

QUESTION: 2

Differentiate with respect to x.

Solution:

y = e^(-x)^{2}.................(1)

Put u = (-x)^{2}

du/dx = -2x dx

Differentiating eq(1) y = e^{u}

dy/du = e^u

⇒ dy/dx = (dy/du) * (du/dx)

= (e^{u}) * (-2x)

⇒ - 2xe^{(-x2)}

QUESTION: 3

The derivatve of f(x) =

Solution:

f(x) = e^{log(logx3)}

As log and e are inverse operations, therefore

f(x)= logx^{3}

So derivative will be {Derivative of log x is 1/x and then applying chain rule}

f’(x)=1/x^{3}. 3x^{2 }= 3/x

QUESTION: 4

The sum of Which series is denoted by e.

Solution:

QUESTION: 5

Derivatve of f(x) is given by

Solution:

y

The derivative of y = e^{f(x)}is dy/dx = f'(x)e^{f(x)}

In this case, f(x) = x^{2 }, and the derivative of x^{2 }= 2x

Therefore, f'(x)= 2x,

dy/dx = 2x

QUESTION: 6

The differential equation of y =

Solution:

QUESTION: 7

If f(x) = e^{2x-5}, then f'(x) is

Solution:

QUESTION: 8

Whta is the derivatve of y = log_{5} (x)

Solution:

y = log5 x = ln x/ln 5 → change of base

= ln x/ln 5

dy/dx = 1/ln5⋅1/x → 1/ln5 is a constant, so we don't change it

= 1/(x ln 5)

QUESTION: 9

For what values of x is e^{logx} = x

Solution:

QUESTION: 10

If f(x) = e^{x}, then the value of f'(-3) is

Solution:

As, f (x) = e^{x}

Similarly, f (-3) = e^{-3}

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