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

Question

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.

Answer 1

$$\lim_{h\rightarrow0}\frac{\sqrt{(x+h+1)^3}-\sqrt{(x+1)^3}}{h}=\lim_{h\rightarrow0}\frac{(x+h+1)^3-(x+1)^3}{h\left(\sqrt{(x+h+1)^3}+\sqrt{(x+1)^3}\right)}=$$ $$=\lim_{h\rightarrow0}\frac{3h(x+1)^2+3h^2(x+1)+h^3}{h\left(\sqrt{(x+h+1)^3}+\sqrt{(x+1)^3}\right)}=\frac{3(x+1)^2}{\sqrt{(x+1)^3}+\sqrt{(x+1)^3}}=\frac{3}{2}\sqrt{x+1}.$$

Answer 2

Let $f(x)=(x+1)^{3/2}$, then $$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h} \cdot \frac {(x+1+h)^{3/2}+(x+1)^{3/2}}{(x+1+h)^{3/2}+(x+1)^{3/2}}= \lim_{h \to 0} \frac{(x+1+h)^3 -(x+1)^3}{h[(x+1+h)^{3/2}+(x+1)^{3/2}]} = \lim_{h \to 0} \frac{h[h^2 + 3hx + 3h + 3x^2 + 6x + 3]}{h[(x+1+h)^{3/2}+(x+1)^{3/2}]} = \lim_{h \to 0} \frac{h^2 + 3hx + 3h + 3x^2 + 6x + 3}{(x+1+h)^{3/2}+(x+1)^{3/2}} = \frac{3x^2+6x+3}{2(x+1)^{3/2}} = \frac{3}{2} \frac{(x+1)^2}{(x+1)^{3/2}}=\frac{3}{2}(x+1)^{1/2}$$

which is exactly what we get from the chain rule:

$$\frac d{dx}(x+1)^{3/2}=\frac{3}{2}(x+1)^{1/2} \cdot\frac{d}{dx}(x)=\frac 32 \sqrt{x+1}.$$ 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.