Differentiation Derivatives of composite Functions
composite Functions (Function of another Function):
Theorem-If `y=f(u)` is differentiable function of u and `u=g(x)` is differentiable function of x such that the composite function `y=f[g(x)]` is differentiable function of x then `\frac{dy}{dx}=\frac{du}{dx}\times \frac{du}{dx}`
1.The derivative of a composite function can also be expressed as follows `y=f(u)` is differentiable function of u and `u =g(x)` is differentiable function of x such that the composite function
`y=f[g(x)]`= `f^{'}[(x)]g^{'}(x)`
2.If `y=f(v)` 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{dy}{dx}=\frac{dy}{dv}\times\frac{dv}{du}\times\frac{du}{dx}`
3. If `y=f(v)` is diffrentiable function `u_{1}`,`u_{i}` is differentiable function for `u_{i+1}` for `i=1,2,..,n-1` and `u_{n}` is diffferentiable function of x, then
`\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}`
`\frac{dy}{dx}`=`\frac{1}{2\sqrt{x^2+5}``\frac{d}{dx}``x^{2}`
` =\frac{1}{2\sqrt{x^2+5}``\times 2x`
`=\frac{x}{\sqrt{x^2+5}}`
(2) `y=sin(log x)`
`\frac{dy}{dx}=sin(log x)``\frac{d}{dx}`(log x)
`=cos(logx) \times \frac{1}{x}`
`=\frac{cos(log x)}{x}`
(3)
`\frac{dy}{dx}`
=`e^{tanx}``\frac{d}{dx}(tan x)`
=`e^{tanx}\times sec^{2}(x)`
(4)`y=log(x^{5}+4)`
`\frac{dy}{dx}`
=`\frac{1}{log (x^{5}+4)}``\frac{d}{dx}x^{5}+4`
=`\frac{1}{log (x^{5}+4)}``\times 5x^{3}+4`
=`\frac{5x^{3}+4}{log (x^{5}+4)}`
(5)
`5^{3cosx-2}`
`\frac{dy}{dx}`
=`5^{3cosx-2}`` \times`` \frac{d}{dx}``(3cosx-2)`
`=5^{3cosx-2}`` \times`` -3cosx`
(6)
y=`\frac{3}{(2x^2-7)^5}`
`y=\frac{3}{{(2x^2-7)}^5}`
`=3 \times \frac{-1}{{(2x^2-7)}^5}^2}``\frac{d}{dx}``{(2x^2-7)}^5`
`=3 \times \frac{-1}{{(2x^2-7)}^10}``5{(2x^2-7)}^4``\frac{d}{dx}``(2x^2-7)`
`=\frac{-3\times5{(2x^2-7)}^4}4x`
Method 2
(1) `y=\sqrt{x^2+5}`
Answer- let `u=x^2+5` then `y=\sqrt{u}` where y is a diffferentiable function of u is a diffferentiable function of x then `\frac{dy}{dx}`
=`\frac{dy}{du}\times\frac{du}{dx}`
differentiate w.r.t u
`\frac{dy}{du}`
=`\frac{1}{2\sqrt{u}}``\frac{du}{dx}`
=`\frac{d}{dx}(x^2+5)`
 |
| Differentiation Derivatives of composite Functions |
