Find the derivatives to $f(x)=4/x^2$ and $g(t)=(t-5)/(1+\sqrt{t}\,)$ I have these two assignments: 

Find the derivatives to (a) $f(x)=4/x^2$ and b) $g(t)=(t-5)/(1+\sqrt{t}\,)$ by using the definition $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$$

a) This is my attempt at (a); am I correct?
$$\lim_{h \to 0} \frac{\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)$$
Then I found the common denominator 
$$\begin{align*}
\frac{4}{(x+h)^2}-\frac{4}{x^2}
&=\frac{4x^2}{(x+h)^2x^2}-\frac{4(x+h)^2}{(x+h)^2x^2} \\[6pt]
&=\frac{4x^2-4x^2-4h^2-8xh}{(x+h)^2x^2} \\[6pt]
&=\frac{-4h^2-8xh}{(x+h)^2x^2}
\end{align*}$$
Then I expand the denominator to
$$\frac{-4h^2-8xh}{x^4+2x^3h+h^2x^2}
=\frac{\left(-4h^2-8h\right)}{x^3+2x^3h+h^2x^2}\frac{1}{h}
=\frac{\left(-4h^2-8h\right)}{hx^3+2x^3h+h^2x^2}$$
And $h$ goes out with:
$$\frac{-4h^2-8}{x^3+2x^3h+h^2x^2}
= \lim_{h \to 0}\frac{-4\cdot0^2-8}{x^3+2x^3\cdot 0+0^2\cdot x^2}
= -\frac{8}{x^3}$$
I dont know how to do part (b)
 A: *

*First function


$$f'(x)=\lim_{h \to 0} \frac{\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)=\lim_{h \to 0}\frac{1}{h}\frac{-4h^2-8xh}{(x+h)^2x^2}=\lim_{h \to 0}\frac{1}{h}\frac{-4h(h+2x)}{(x+h)^2x^2}= \lim_{h \to 0}\frac{-4(h+2x)}{(x+h)^2x^2}= \frac{-8x}{x^4}= -\frac{8}{x^3}, $$
for all $x\neq 0$.


*Second function (in what follows we consider $t\geq 0$)


$$f'(t)=\lim_{h \to 0} \frac{1}{h}\left(\frac{t+h-5}{1+\sqrt{t+h}}-\frac{t-5}{1+\sqrt{t}}
\right) = 
\lim_{h \to 0} \frac{1}{h}\left(\frac{(t+h-5)(1+\sqrt{t})-(t-5)(1+\sqrt{t+h})}{(1+\sqrt{t+h})(1+\sqrt{t})}\right);$$
Now we simplify the numerator  arriving at
$$f'(t)=
 \lim_{h \to 0}
\frac{1}{h}\left(\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})+h(1+\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}\right)= \lim_{h \to 0} \frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}+\frac{1}{h}\frac{h(1+\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}= 
\lim_{h \to 0} \frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}+\frac{1}{(1+\sqrt{t+h})};
$$
All we need is to use the "trick":
$$\frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}=
\frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}\frac{(\sqrt{t+h}+\sqrt{t})}{(\sqrt{t+h}+\sqrt{t})}=
\frac{1}{h}\frac{-(t-5)h}{(1+\sqrt{t+h})(1+\sqrt{t})}\frac{1}{(\sqrt{t}+\sqrt{t+h})}=
-\frac{(t-5)}{(1+\sqrt{t+h})(1+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}.
$$
In summary:
$$f'(t)= \lim_{h \to 0} -\frac{(t-5)}{(1+\sqrt{t+h})(1+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}+\frac{1}{(1+\sqrt{t+h})}= \\
-\frac{(t-5)}{2(1+\sqrt{t})^2\sqrt{t}}+\frac{1}{(1+\sqrt{t})}. $$ 
