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
Which class you are presently in?
Choose an option. You can change your option at any time.
You will be solving questions and growing your critical thinking skills.   