Question

Determine f’(0) for f(x) = 2 sin x cos x

Answer 1

Given:

`f(x)=2sinx.cosx`

To find: Determine f'(0)=?

Step-by-step explanation:

We have given that

`f(x)=2sinx.cosx.......(1)`

We know the identity of the trigonometric

`sin2x=2sinx.cosx`

Eq.1 becomes

`f(x)=sin2x`

Now, differentiate f(x) w.r.to x

Therefore,

`\begin{gathered} f^(\prime)(x)=(d)/(dx)[sin(2x)] \\ \\ f^(\prime)(x)=cos2x*(d)/(dx)(2x) \\ \\ f^(\prime)(x)=cos2x*(2) \\ \\ f^(\prime)(x)=2cos2x \end{gathered}`

Hence,

`f^(\prime)(x)=2cos2x`
`f^(\prime)(x)=2cos2x`

For the value of f'(0) put x = 0 in f'(x)

We get,

`\begin{gathered} f^(\prime)(0)=2cos2(0) \\ \\ f^(\prime)(0)=2* cos0 \end{gathered}`

We know that

`cos0=1`

So,

`\begin{gathered} f^(\prime)(0)=2*1 \\ \\ f^(\prime)(0)=2 \end{gathered}`

Answer: f'(0) = 2 for f(x) = 2 sin x cos x.