Differentiation Derivatives of composite Functions

composite Functions (Function of another Function):

Theorem-If $y = f (u)$ $y=f(u)$ is differentiable function of u and $u = g (x)$ $u=g(x)$ is differentiable function of x such that the composite function $y = f [g (x)]$ $y=f[g(x)]$ is differentiable function of x then $\frac{d y}{d x} = \frac{d u}{d x} \times \frac{d u}{d x}$ $\frac{dy}{dx}=\frac{du}{dx}\times \frac{du}{dx}$

Differentiation Derivatives of composite Functions
Differentiation Derivatives of composite Functions
Differentiate the following w.r.t x :
(1) $y = \sqrt{x^{2} + 5}$ $y=\sqrt{x^2+5}$

Note-

1.The derivative of a composite function can also be expressed as follows $y = f (u)$ $y=f(u)$ is differentiable function of u and $u = g (x)$ $u =g(x)$ is differentiable function of x such that the composite function

$y = f [g (x)]$ $y=f[g(x)]$ = $f^{'} [(x)] g^{'} (x)$ $f^{'}[(x)]g^{'}(x)$

2.If `y=f(v) $i s d \Leftrightarrow e r e n t i a b \leq f u n c t i o n o f v and$ $is differentiable function of v and$ v=g(u)` is diffferentiable function of u and u=h(x) is a differentiable function of x then

$\frac{d y}{d x} = \frac{d y}{d v} \times \frac{d v}{d u} \times \frac{d u}{d x}$ $\frac{dy}{dx}=\frac{dy}{dv}\times\frac{dv}{du}\times\frac{du}{dx}$

3. If $y = f (v)$ $y=f(v)$ is diffrentiable function $u_{1}$ $u_{1}$ , $u_{i}$ $u_{i}$ is differentiable function for $u_{i + 1}$ $u_{i+1}$ for $i = 1, 2, . ., n - 1$ $i=1,2,..,n-1$ and $u_{n}$ $u_{n}$ is diffferentiable function of x, then

$\frac{d y}{d x} = \frac{d y}{d u_{1}} \times \frac{d u_{1}}{d u_{2}} \times ... . \times \frac{d u_{n - 1}}{d u_{n}} \times \frac{d u_{n}}{d x}$ $\frac{dy}{dx}=\frac{dy}{du_{1}}\times \frac{du_{1}}{du_{2}}\times....\times\frac{du_{n-1}}{du_n}\times\frac{du_{n}}{dx}$

this rule is called chain rule.

Method 2

Differentiation Derivatives of composite Functions

Differentiate the following w.r.t x :

(1) $y = \sqrt{x^{2} + 5}$ $y=\sqrt{x^2+5}$

$\frac{d y}{d x}$ $\frac{dy}{dx}$ = $\frac{1}{2 \sqrt{x^{2} + 5}}$ $\frac{1}{2\sqrt{x^2+5}$ $\frac{d}{d x}$ $\frac{d}{dx}$ $x^{2}$ $x^{2}$

$= \frac{1}{2 \sqrt{x^{2} + 5}}$ $=\frac{1}{2\sqrt{x^2+5}$ $\times 2 x$ $\times 2x$

$= \frac{x}{\sqrt{x^{2} + 5}}$ $=\frac{x}{\sqrt{x^2+5}}$

(2) `y=sin(log x)`

$\frac{d y}{d x} = \sin (\log x)$ $\frac{dy}{dx}=sin(log x)$ $\frac{d}{d x}$ $\frac{d}{dx}$ (log x)

$= \cos (\log x) \times \frac{1}{x}$ $=cos(logx) \times \frac{1}{x}$

$= \frac{\cos (\log x)}{x}$ $=\frac{cos(log x)}{x}$

(3)

$y=e^{tanx}$

$\frac{d y}{d x}$ $\frac{dy}{dx}$

= $e^{\tan x}$ $e^{tanx}$ $\frac{d}{d x} (\tan x)$ $\frac{d}{dx}(tan x)$

= $e^{\tan x} \times \sec^{2} (x)$ $e^{tanx}\times sec^{2}(x)$

(4) $y = \log (x^{5} + 4)$ $y=log(x^{5}+4)$

$\frac{d y}{d x}$ $\frac{dy}{dx}$

= $\frac{1}{\log (x^{5} + 4)}$ $\frac{1}{log (x^{5}+4)}$ $\frac{d}{d x} x^{5} + 4$ $\frac{d}{dx}x^{5}+4$

= $\frac{1}{\log (x^{5} + 4)}$ $\frac{1}{log (x^{5}+4)}$ $\times 5 x^{3} + 4$ $\times 5x^{3}+4$

= $\frac{5 x^{3} + 4}{\log (x^{5} + 4)}$ $\frac{5x^{3}+4}{log (x^{5}+4)}$

(5)

$5^{3 \cos x - 2}$ $5^{3cosx-2}$

5^{3 \cos x - 2}

$5^{3cosx-2}$

$\frac{d y}{d x}$ $\frac{dy}{dx}$

= $5^{3 \cos x - 2}$ $5^{3cosx-2}$ $\times$ $\times$ $\frac{d}{d x}$ $\frac{d}{dx}$ $(3 \cos x - 2)$ $(3cosx-2)$

$= 5^{3 \cos x - 2}$ $=5^{3cosx-2}$ $\times$ $\times$ $- 3 \cos x$ $-3cosx$

(6)

y= $\frac{3}{{(2 x^{2} - 7)}^{5}}$ $\frac{3}{(2x^2-7)^5}$

$y = \frac{3}{{(2 x^{2} - 7)}^{5}}$ $y=\frac{3}{{(2x^2-7)}^5}$

$= 3 \times {\frac{- 1}{{(2 x^{2} - 7)}^{5}}}^{2}}$ $=3 \times \frac{-1}{{(2x^2-7)}^5}^2}$ $\frac{d}{d x}$ $\frac{d}{dx}$ ${(2 x^{2} - 7)}^{5}$ ${(2x^2-7)}^5$

$= 3 \times \frac{- 1}{{(2 x^{2} - 7)}^{10}}$ $=3 \times \frac{-1}{{(2x^2-7)}^10}$ $5 {(2 x^{2} - 7)}^{4}$ $5{(2x^2-7)}^4$ $\frac{d}{d x}$ $\frac{d}{dx}$ $(2 x^{2} - 7)$ $(2x^2-7)$

$= \frac{- 3 \times 5 {(2 x^{2} - 7)}^{4}}{4} x$ $=\frac{-3\times5{(2x^2-7)}^4}4x$

Method 2

(1) $y = \sqrt{x^{2} + 5}$ $y=\sqrt{x^2+5}$

Answer- let $u = x^{2} + 5$ $u=x^2+5$ then $y = \sqrt{u}$ $y=\sqrt{u}$ where y is a diffferentiable function of u is a diffferentiable function of x then $\frac{d y}{d x}$ $\frac{dy}{dx}$

= $\frac{d y}{d u} \times \frac{d u}{d x}$ $\frac{dy}{du}\times\frac{du}{dx}$

differentiate w.r.t u

$\frac{d y}{d u}$ $\frac{dy}{du}$

= $\frac{1}{2 \sqrt{u}}$ $\frac{1}{2\sqrt{u}}$ $\frac{d u}{d x}$ $\frac{du}{dx}$

= $\frac{d}{d x} (x^{2} + 5)$ $\frac{d}{dx}(x^2+5)$

Differentiation Derivatives of composite Functions

Differentiation Derivatives of composite Functions

Differentiation Derivatives of composite Functions

Differentiation Derivatives of composite Functions

Differentiate the following w.r.t x :

(1) $y = \sqrt{x^{2} + 5}$ $y=\sqrt{x^2+5}$

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Differentiation Derivatives of composite Functions

Differentiation Derivatives of composite Functions

Differentiation Derivatives of composite Functions

Differentiate the following w.r.t x :

(1) y=√x2+5y=x2+5y=\sqrt{x^2+5}

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Books, Practicles

Maharashtra Board & All Books

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Circular Permutation 11th Class Maharashtra Board

Numerical Taxonomy (Principles Example) Cladogram Phylogeny DNA barcoding Biology

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(1) $y = \sqrt{x^{2} + 5}$ $y=\sqrt{x^2+5}$