Differentiate using the definition of the derivative?

**magnifico** · Oct 23rd 2009, 09:33 AM

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

**Amer** · Oct 23rd 2009, 09:53 AM

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

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

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

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

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

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

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

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