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. 
$$\lim\limits_{h \to 0 } \frac{\sin^2\left(\frac\pi 4+h \right)-\frac 1 2} h$$

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}
