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The chain rule is used to find the derivatives of inbuilt functions such as cos(sinx) here sinx is the inner function of a function so let us try to evaluate this type of functions using derivative chain rule.^{}

Let p = g(x) be derivable at ‘x’ and y = f(p) be derivable at the corresponding value of (p). Then the composite function y = f(g(x)) is derivable at x and

$(\frac{dy}{dx})=(\frac{dy}{dp}).(\frac{dp}{dx})$

The differentiation of ‘y’ with respect to ‘x’ is equal to the differentiation of ‘y’ with respect to ‘p’ times the differentiation of ‘p’ with respect to ‘x’.

Let $\Delta$ p and $\Delta$ y be the increments of ‘p’ and ‘y’ respectively corresponding to the increment $\Delta$ x of ‘x’. The relation

$\lim_{_{\Delta p\rightarrow 0}}\frac{\Delta y}{\Delta u}=\frac{dy}{dp}$

Limit p tends to ‘0’ ($\Delta$y /$\Delta$p) is equal to differentiation of ‘y’ with respect ‘p’

Implies that

$\frac{\Delta y}{\Delta p}=\frac{dy}{dp}+\varepsilon$ , where $\varepsilon \rightarrow 0$ as $\Delta p\rightarrow 0$

$\Delta y=\frac{dy}{dp}.\Delta p+\varepsilon .\Delta p$

$\frac{\Delta y}{\Delta x}=\frac{dy}{dp}.\frac{\Delta p}{\Delta x}+\varepsilon .\frac{\Delta p}{\Delta x}......................(i)$

Since ‘p’ is derivable at ‘x’, it is continuous at x and so

$\lim_{_{\Delta x\rightarrow 0}}\Delta p=0$

Hence $\varepsilon \rightarrow 0$ as $\Delta x\rightarrow 0$

Taking the limit as $\Delta x\rightarrow 0$ in (i), we obtain

$\frac{dy}{dx}=\frac{dy}{dp}.\frac{dp}{dx}+0.\frac{dp}{dx}$

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

Hence, the theorem is proved.

⇒ Let us understand the result of the above theorem with the following derivative chain rule example

Let y = 10x + 4

y = 2(5x + 2) - - - - - - - (i)

Here we first recognize that ‘y’ is a composite function and then we need to split up the composite function. In this case, the inside function is p = 5x + 2 - - - - - - (ii)

And the outside function is y = 2p - - - - - - - (iii)

Now, differentiating (i) with respect to ‘x’ we get $\frac{dy}{dx}$ = 10

Differentiating (ii) with respect to ‘x’ we get, $\frac{dp}{dx}$ - - - - - - (iv)

Differentiating (iii) with respect to ‘u’ we get, $\frac{dy}{dp}$ = 2 - - - - - - - - (v)

$(\frac{dy}{dp})*(\frac{dp}{dx})=5*2=10............(vi)$

From (i) and (vi) we get

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

Find $(\frac{d^{2}p}{dx^{2}})$ if p = f(u, v) = uv where u = 2cos (x) and v = 2sin (x)

We first find the first derivative with respect to ‘x’ and then we find the second derivative.

Given u = 2cos (x) and v = 2sin (x)

We know

$(\frac{dp}{dx})$ = - 4 sin

Now, differentiating $(\frac{dp}{dx})$ with respect to ‘x’ we get

$(\frac{d^{2}p}{dx^{2}})=D[(\frac{dp}{dx})]$

$(\frac{d^{2}p}{dx^{2}})$ = D[- 4 sin

$(\frac{d^{2}p}{dx^{2}})$ = [- 4 * 2sin(x) cos(x) + 4 * - 2 cos(x) sin(x)]

$(\frac{d^{2}p}{dx^{2}})$ = [- 8sin(x) cos(x) - 8 cos(x) sin(x)]

$(\frac{d^{2}p}{dx^{2}})$ = - 16sin(x) cos(x)

$(\frac{d^{2}p}{dx^{2}})$ = - 8 sin(2x)