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Basic Differentiation Rule

Find the deri...

Question

Find the derivative of $\frac{1}{a x^{2} + b x + c}$ where $a, b, c$ are fixed non-zero constants.

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Solution

Let $f (x) = \frac{1}{a x^{2} + b x + c}$
Differentiating with respect to $x$
$\Rightarrow f^{'} (x) = \frac{(\frac{d}{d x} (1)) (a x^{2} + b x + c) - (\frac{d}{d x} (a x^{2} + b x + c)) (1)}{(a x^{2} + b x + c)^{2}}$
$\Rightarrow f^{'} (x) = \frac{(0) (a x^{2} + b x + c) - (2 a x + b) (1)}{(a x^{2} + b x + c)^{2}}$
$\Rightarrow f^{'} (x) = \frac{0 - (2 a x + b)}{(a x^{2} + b x + c)^{2}}$
$∴ f^{'} (x) = \frac{- (2 a x + b)}{(a x^{2} + b x + c)^{2}}$

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Basic Differentiation Rule

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