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Fundamental rules for differentiation

 

In the previous article we have used the definition of derivatives to find derivatives.This section is mainly devoted several rules that allows us to find derivatives without using definition directly.

 

  1. Rule (I): Differentiation of constant function is zero i.e., $latex \displaystyle \frac { d }{ dx } \left( c \right) =0 &s=1$
  2. Rule (II): Let f(x) be a differentiable function and let c be a constant. Then c.f(x) is also differentiable such that
    $latex \displaystyle \frac { d }{ dx } \left( c.f\left( x \right) \right) =c.\frac { d }{ dx } \left( f\left( x \right) \right) &s=1 $
  3. Rule (III): If f(x) and g(x) are differentiable functions, then show that $latex \displaystyle f\left( x \right)\pm g\left( x \right) &s=1$ are also differentiable such that
    $latex \displaystyle \frac { d }{ dx } \left[ f\left( x \right) \pm g\left( x \right) \right] =\frac { d }{ dx } f\left( x \right) \pm \frac { d }{ dx } g\left( x \right) &s=1 $
  4. Rule (IV) (Product Rule): If f(x) and g(x) are two differentiable functions, then
    $latex \displaystyle \frac { d }{ dx } \left[ f\left( x \right) .g\left( x \right) \right] =f\left( x \right) \frac { d }{ dx } \left\{ g\left( x \right) \right\} +\frac { d }{ dx } \left\{ f\left( x \right) \right\} g\left( x \right) &s=1$
  5. Rule (v) (Quotient Rule): If f(x) and g(x) are two differentiable functions and $latex \displaystyle g\left( x \right) \neq 0\quad then\\ \frac { d }{ dx } \left\{ \frac { f\left( x \right) }{ g\left( x \right) } \right\} =\frac { g\left( x \right) \frac { d }{ dx } \left\{ f\left( x \right) \right\} -f\left( x \right) .\frac { d }{ dx } \left\{ g\left( x \right) \right\} }{ { \left[ g\left( x \right) \right] }^{ 2 } } &s=1 $
April 18, 2019
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