Question

If f(sinx)=(tanx)^2 then find f'(x)
Ans:- 2x/(1-x^2)^2
plz sopve this..​

Answer 1

ANSWER :–

f'(x) = 2x/(1 - x²)²

EXPLANATION :–

GIVEN :–

A function f(sinx)=(tanx)² .

TO FIND:–

Value of f'(x)

SOLUTION :–

• Given function –

=> f(sinx) = (tanx)²

• Now let's put sin(x) = x –

• And we know that –

=> cos(x) = √[1 - sin²(x)]

=> cos(x) = √(1 - x²)

• Now tan(x) = sin(x)/cos(x) –

=> tan(x) = x/√(1 - x²)

• Now put the values in function –

=> f(x) = [x/√(1 - x²)]²

=> f(x) = x²/(1 - x²)

Differentiate with respect to 'x' –

=> f'(x) = [(1 - x²)(2x) - x²(-2x)] / [(1 - x²)²]

=> f'(x) = [(1 - x²)(2x) + x²(2x)] / [(1 - x²)²]

=> f'(x) = [(1 - x² + x²)(2x)] / [(1 - x²)²]

=> f'(x) = 2x/(1 - x²)²

USED FORMULA :–

(1) If f(x) = U/V , then f'(x) = [VU' - UV']/V²