Thread: Differentiate using the definition of the derivative?

G(x) = (1)/(√(2x + 3))
Differentiate using the definition of the derivative.
The answer should be: G′ (x) = (−1)/((2x + 3)^3/2)
When you plug it into f(x+h)-f(x)/h, I think you multiply by the conjugate but I'm not entirely sure... Can someone explain how to get the answer. Explanations of the steps used would be nice too

Originally Posted by magnifico

G(x) = (1)/(√(2x + 3))
Differentiate using the definition of the derivative.
The answer should be: G′ (x) = (−1)/((2x + 3)^3/2)
When you plug it into f(x+h)-f(x)/h, I think you multiply by the conjugate but I'm not entirely sure... Can someone explain how to get the answer. Explanations of the steps used would be nice too

$\displaystyle \lim _{y\rightarrow x} \frac{f(y)-f(x)}{y-x}$

$\displaystyle \lim_{y\rightarrow x} \dfrac{\dfrac{1}{\sqrt{2y+3}} - \dfrac{1}{\sqrt{2x+3}}}{y-x} $

$\displaystyle \lim _{y\rightarrow x} \frac{\sqrt{2x+3}-\sqrt{2y+3}}{(y-x)\sqrt{2x+3}\sqrt{2y+3}}$

$\displaystyle \lim _{y\rightarrow x} \frac{\sqrt{2x+3}-\sqrt{2y+3}}{(y-x)\sqrt{2x+3}\sqrt{2y+3}}\times \frac{\sqrt{2x+3}+\sqrt{2y+3}}{\sqrt{2x+3}+\sqrt{2 y+3}}$

$\displaystyle \lim _{y\rightarrow x} \frac{2x-2y}{(y-x)\sqrt{2x+3}\sqrt{2y+3}}\times \frac{1}{\sqrt{2x+3}+\sqrt{2y+3}}$

$\displaystyle \lim _{y\rightarrow x} \frac{-2}{\sqrt{2x+3}\sqrt{2y+3}}\times \frac{1}{\sqrt{2x+3}+\sqrt{2y+3}}=\frac{-1}{(2x+3)\sqrt{2x+3}}$