Given:
[tex]f(x)=2sinx.cosx[/tex]To find: Determine f'(0)=?
Explanation:
We have given that
[tex]f(x)=2sinx.cosx.......(1)[/tex]We know the identity of the trigonometric
[tex]sin2x=2sinx.cosx[/tex]Eq.1 becomes
[tex]f(x)=sin2x[/tex]Now, differentiate f(x) w.r.to x
Therefore,
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}[sin(2x)] \\ \\ f^{\prime}(x)=cos2x\times\frac{d}{dx}(2x) \\ \\ f^{\prime}(x)=cos2x\times(2) \\ \\ f^{\prime}(x)=2cos2x \end{gathered}[/tex]Hence,
[tex]f^{\prime}(x)=2cos2x[/tex][tex]f^{\prime}(x)=2cos2x[/tex]For the value of f'(0) put x = 0 in f'(x)
We get,
[tex]\begin{gathered} f^{\prime}(0)=2cos2(0) \\ \\ f^{\prime}(0)=2\times cos0 \end{gathered}[/tex]We know that
[tex]cos0=1[/tex]So,
[tex]\begin{gathered} f^{\prime}(0)=2\times1 \\ \\ f^{\prime}(0)=2 \end{gathered}[/tex]Answer: f'(0) = 2 for f(x) = 2 sin x cos x.
