Question

f(x)=sin^2-cos^2 then find the value of f'(π/4)

Answer 1

Hey there !!!!

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f(x) = sin²x-cos²x

sin²x+cos²x=1

So,

f(x)= 1-cos²x-cos²x

f(x)=1-2cos²x

But cos2x= 2cos²x-1

f(x)= -(2cos²x-1)

f(x)= -cos2x

derivative of cost  = -sint*dt

f¹(x) = -2*(-sin2x)

f¹(x) =2sin2x

f¹(π/4) =2sin2*π/4

f¹(π/2) =2sinπ/2

sinπ/2=1

So,

f¹(π/4) =2*1=2

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Hope this helped you.............

Answer 2

Answer:

2

Explanation:

fx = -(cosx - sinx)

fx = -Cos2x

f'x = -(-sin2x) . d( 2x)/dx

f'x = sin2x . 2 = 2sin2x

 π /4 = 45⁰

f'(45⁰) = sin ( 2×45⁰) . 2

= sin 90⁰ × 2

= 1×2 = 2