Ucale

# Rule for Differentiation of a function

### Chain Rule:   If f(x) and g(x) are differentiable functions, then f(g(x)) is also differentiable

$\displaystyle f\left( g\left( x \right) \right) =\frac { d }{ dg\left( x \right) } f\left( g\left( x \right) \right) \times \frac { d }{ dx } g\left( x \right) \\ i.e.,\quad If\quad z=f\left( x \right) \quad and\quad y=g\left( x \right) ,\quad then\\ \frac { dz }{ dx } =\frac { dz }{ dx } .\frac { dy }{ dx }$

### Example Differentiate w.r.t. x

$\displaystyle \sin { \left( { x }^{ 2 }+1 \right) } \\ \Rightarrow Let\quad y=\sin { \left( { x }^{ 2 }+1 \right) } \\ Put\quad u={ x }^{ 2 }+1\\ So,\quad y=\sin { u } \\ and\quad \frac { dy }{ dx } =\frac { dy }{ du } .\frac { du }{ dx } \quad \\ \Rightarrow \frac { dy }{ du } =\cos { u } \quad and\quad \frac { du }{ dx } =2x\\ so,\frac { dy }{ dx } =\cos { u } .2x\\ \frac { dy }{ dx } =2x\cos { \left( { x }^{ 2 }+1 \right) } \qquad \left[ \because u={ x }^{ 2 }+1 \right] \\ \therefore \frac { d }{ dx } \left( \sin { \left( { x }^{ 2 }+1 \right) } \right) =2x\cos { \left( { x }^{ 2 }+1 \right) }$

April 18, 2019
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