# How to differentiate $(x+1)^{3/2}$ using the limit definition of the derivative?

I want to use the definition of the derivative to find $f'$ of $f(x)=(x+1)^{3/2}$.

I solved it using the chain rule. Would like to try to solve it using the definition of a derivative:

$$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.$$

Don't really know how I should start it. Should I make it $\sqrt{(x+1)^3}$? Or keep the exponent as $3/2$? Other guidance would be appreciated.

• If you know how the chain rule is proven, then you can follow those steps to make the definition work – Kaynex Oct 14 '17 at 4:38

$$\lim_{h\rightarrow0}\frac{\sqrt{(x+h+1)^3}-\sqrt{(x+1)^3}}{h}=\lim_{h\rightarrow0}\frac{(x+h+1)^3-(x+1)^3}{h\left(\sqrt{(x+h+1)^3}+\sqrt{(x+1)^3}\right)}=$$ $$=\lim_{h\rightarrow0}\frac{3h(x+1)^2+3h^2(x+1)+h^3}{h\left(\sqrt{(x+h+1)^3}+\sqrt{(x+1)^3}\right)}=\frac{3(x+1)^2}{\sqrt{(x+1)^3}+\sqrt{(x+1)^3}}=\frac{3}{2}\sqrt{x+1}.$$
Let $f(x)=(x+1)^{3/2}$, then $$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h} \cdot \frac {(x+1+h)^{3/2}+(x+1)^{3/2}}{(x+1+h)^{3/2}+(x+1)^{3/2}}= \lim_{h \to 0} \frac{(x+1+h)^3 -(x+1)^3}{h[(x+1+h)^{3/2}+(x+1)^{3/2}]} = \lim_{h \to 0} \frac{h[h^2 + 3hx + 3h + 3x^2 + 6x + 3]}{h[(x+1+h)^{3/2}+(x+1)^{3/2}]} = \lim_{h \to 0} \frac{h^2 + 3hx + 3h + 3x^2 + 6x + 3}{(x+1+h)^{3/2}+(x+1)^{3/2}} = \frac{3x^2+6x+3}{2(x+1)^{3/2}} = \frac{3}{2} \frac{(x+1)^2}{(x+1)^{3/2}}=\frac{3}{2}(x+1)^{1/2}$$
$$\frac d{dx}(x+1)^{3/2}=\frac{3}{2}(x+1)^{1/2} \cdot\frac{d}{dx}(x)=\frac 32 \sqrt{x+1}.$$ The key step is in the third equality above, where you multiply the numerator and denominator by the conjugate to get rid of the square root.
You can also use the product rule: $$\left[(x+1)^{3/2}\right]'=\left[(x+1)\sqrt{x+1}\right]'=$$ $$\sqrt{x+1}\cdot\lim_{h\rightarrow 0} \frac{(x+h+1)-(x+1)}{h}+(x+1)\cdot\lim_{h\rightarrow 0} \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}=$$ $$\sqrt{x+1}+(x+1)\cdot \lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h+1}+\sqrt{x+1})}=$$ $$\sqrt{x+1}+\frac{\sqrt{x+1}}{2}=\frac{3}{2}\sqrt{x+1}.$$