Question

f(x)=sin2x-cos2x then the value of f'(π/4)​

Answer 1

Answer:

2

Step-by-step explanation:

As, f'(x) =2cos2x+2sin2x

And at x=¶/4

F'(¶/4)=2cos¶/2+2sin¶/2

=0+2(1)=2

Thank you

Answer 2

Answer:

The value of f'(π/4) is 2.

Step-by-step explanation:

Given,

f(x) = sin(2x) - cos(2x)

To find,

The value of f'(π/4)

Calculation,

f(x) = sin(2x) - cos(2x) ...(1)

Now we differentiate equation (1) with respect to x once.

⇒ f'(x) = cos(2x) (2) - (- sin(2x)(2))

Properties used are:

• (sinx)' = cosx
• (cosx)' = -sinx
• f'(kx) = kf(x)

Hence,

f'(x) = 2cos(2x) + 2sin(2x)

Now the value of f'(x) at x = π/4 is:

f'(π/4) = 2cos(2(π/4)) + 2sin(2(π/4))

⇒ f'(π/4) = 2cos(π/2) + 2sin(π/2)

⇒ f'(π/4) = 2 (As cos(π/2) = 0, and sin(π/2) = 1)

Therefore, the value of f'(π/4) is 2.

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