Chain rule of derivatives Theorem examples with solutions

Chain rule is defined as: The composition $fog$ of function $f$ and $g$ is the function whose value $f[g(x)]$ are found for each $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f.(f[g(x)])$ is read as ( $f$ of $g$ of $x$ )

Theorem

if $g$ is differentiable at point $x$ and $f$ is differentiable at point $g(x)$ then composition function $fog$ is differentiable at point $x$ and $\left(fog\right)'\left(x\right)=f'\left[g\left(x\right)\right].g'\left(x\right)$

Proof:

Let $y=\left(fog\right)\left(x\right)=f\left[g\left(x\right)\right]$

Then $\left(fog\right)'\left(x\right)=\left[f\left[g\left(x\right)\right]\right]'=\frac{dy}{dx}$

$\frac{dy}{dx}=f'\lbrack g\left(x\right)\rbrack.g'(x).....(1)$

Let $u=g(x)................(2)$

Then $y=f(u).............(3)$

Differentiating eq(2) and eq(3) w.r.t $x$ and $u$ respectively

We have $\frac{du}{dx}=\frac d{dx}\left[g\left(x\right)\right]=g'\left(x\right)$

And $\frac{dy}{du}=\frac d{du}\left[f\left(u\right)\right]=f'\left(x\right)$

Thus eq(1) can be written in the following forms

$\frac d{dx}\left[f\left(u\right)\right]=f'\left(u\right)\frac{du}{dx}$

$\boxed{\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}}$

Example 1:  [Chain rule of derivatives]

solve by chain rule $y=(x^3+1)^9$

$y=(x^3+1)^9$ and $u=(x^3+1)$ .

$y=u^6$

Now $\frac{du}{dx}=3x^2\;and\;\frac{dy}{dx}=9u^8$ (Power rule)

Now by using chain Rule $\frac{dy}{dx}=\;\frac{dy}{du}.\frac{du}{dx}$

we have $\frac{dy}{dx}=9u^8\frac{du}{dx}$

or $\frac d{dx}\left(x^3+1\right)^9=9\left(x^3+1\right)^8\left(3x^2\right)$

since $u=x^3+1$ and $\frac{du}{dx}=3x^2$

$\frac d{dx}\left(x^3+1\right)^9=27\left(x^3+1\right)^8$

Hence chain Rule is satisfied

Example 2  [Chain rule of derivatives ]

By using chain rule solve the equation

Let $y=\left(3x^2-2x+1\right)^6$

we take $u=3x^2-2x+7........(1)$

Then $y=u^6...........(2)$

differentiate eq (2) w.r.t $u$

$\frac{dy}{dx}=6u^{6-1}\frac{du}{du}$

$\frac{dy}{dx}=6u^{6-1}\left(1\right)$

$\boxed{\frac{dy}{dx}=6u^5}............(A)$

differentiate eq (1) w.r.t $x$

$\frac{du}{dx}=\frac d{dx}\left(3x^2-2x+7\right)$

$\frac{du}{dx}=\frac d{dx}3x^2-\frac{\displaystyle d}{\displaystyle dx}2x+\frac{\displaystyle d}{\displaystyle dx}7$

$\frac{du}{dx}=3\left(2\right)x-2\left(1\right)+0$

$\boxed{\frac{du}{dx}=6x-2}.........(B)$

Now by using chain rule

$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}$

$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}=6u^5.\left(6x-2\right)$

$\frac{dy}{dx}=6\left(3x^2-2x+7\right)\left(6x-2\right)$

$\boxed{\frac{dy}{dx}=12\left(3x^2-2x+7\right)\left(3x-1\right)}$

Hence chain rule is satisfied.

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