Calculus 1- Find directly the derivative of a function f. The following limit represents the derivative of a function $f$ at a point $a$. Evaluate the limit. 
$$\lim\limits_{h \to 0 } \frac{\sin^2\left(\frac\pi 4+h \right)-\frac 1 2} h$$
 A: Write the limit as
$$\mathop {\lim }\limits_{h \to 0} \frac{{2{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - 1}}{{2h}}$$
Now use  $$2{\sin ^2}x - 1 =  - \cos \left( {2x} \right)$$
Thus
$$=\mathop {\lim }\limits_{h \to 0} \frac{{-\cos \left( {\frac{\pi }{2} + 2h} \right)}}{{2h}}$$
$$ =   \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {2h} \right)}}{{2h}} =   1$$
A: Let $f(x)=\sin^2x$. We have, $f^{\prime}(x)=2\sin x\cos x$. In the other hand
\begin{equation}
\begin{array}{lll}
\lim_{h\rightarrow 0}\frac{\sin^2\left(\frac{\pi}{4}+h\right)-\frac{1}{2}}{h}&=&\lim_{h\rightarrow 0}\frac{f\left(\frac{\pi}{4}+h\right)-f\left(\frac{\pi}{4}\right)}{h}\\
&=&f^{\prime}(\pi/2)\\
&=&2\sin(\pi/4)\cos(\pi/4)\\
&=&2(1/\sqrt{2})(1/\sqrt{2})=1.
\end{array}
\end{equation}
A: I interpret the hint as asking you to identify the function and where the derivative is taken. There are a couple of natural choices. 
If you let $f(x)=\sin^2(\pi/4+x)$, your limit expression, directly from the definition of derivative, is $f'(0)$. Now calculate $f'(x)$ using the ordinary rules of differentiation, and find $f'(0)$.
Or else let $f(x)=\sin^2 x$. Then our limit is the derivative of $f(x)$ at $x=\pi/4$. 
