Question

Find the differential coefficient of

`e^(2x)(1+Lnx)`​

Answer 1

Product Rule for Differentiation

`\textsf{If }y=uv`

`(dy)/(dx)=u(dv)/(dx)+v(du)/(dx)`

Given equation:

`y=e^(2x)(1+\ln x)`

Define the variables:

`\textsf{Let }u=e^(2x) \implies (du)/(dx)=2e^(2x)`

`\textsf{Let }v=1+\ln x \implies (dv)/(dx)=(1)/(x)`

Therefore:

`\begin{aligned}(dy)/(dx) & =u(dv)/(dx)+v(du)/(dx)\\\\\implies (dy)/(dx) & =e^(2x) \cdot (1)/(x)+(1+\ln x) \cdot 2e^(2x)\\\\& = (e^(2x))/(x)+2e^(2x)(1+\ln x)\\\\ & = (e^(2x))/(x)+2e^(2x)+2e^(2x) \ln x\\\\& = e^(2x)\left((1)/(x)+2+2 \ln x \right)\end{aligned}`

Answer 2

`\rm \displaystyle y' = 2 {e}^(2x) + (1)/(x) {e}^(2x) + 2 \ln(x) {e}^(2x)`

Explanation:

we would like to figure out the differential coefficient of
`e^(2x)(1+\ln(x))`

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

`\displaystyle y = {e}^(2x) \cdot (1 + \ln(x) )`

to do so distribute:

`\displaystyle y = {e}^(2x) + \ln(x) \cdot {e}^(2x)`

take derivative in both sides which yields:

`\displaystyle y' = (d)/(dx) ( {e}^(2x) + \ln(x) \cdot {e}^(2x) )`

by sum derivation rule we acquire:

`\rm \displaystyle y' = (d)/(dx) {e}^(2x) + (d)/(dx) \ln(x) \cdot {e}^(2x)`

Part-A: differentiating $e^{2x}$

`\displaystyle (d)/(dx) {e}^(2x)`

the rule of composite function derivation is given by:

`\rm\displaystyle (d)/(dx) f(g(x)) = (d)/(dg) f(g(x)) * (d)/(dx) g(x)`

so let g(x) [2x] be u and transform it:

`\displaystyle (d)/(du) {e}^(u) \cdot (d)/(dx) 2x`

differentiate:

`\displaystyle {e}^(u) \cdot 2`

substitute back:

`\displaystyle \boxed{2{e}^(2x) }`

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

`\displaystyle (d)/(dx) f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)`

let

• 
  `f(x) \implies \ln(x)`
• 
  `g(x) \implies {e}^(2x)`

substitute

`\rm\displaystyle (d)/(dx) \ln(x) \cdot {e}^(2x) = (d)/(dx)( \ln(x) ) {e}^(2x) + \ln(x) (d)/(dx) {e}^(2x)`

differentiate:

`\rm\displaystyle (d)/(dx) \ln(x) \cdot {e}^(2x) = \boxed{(1)/(x) {e}^(2x) + 2\ln(x) {e}^(2x) }`

Final part:

substitute what we got:

`\rm \displaystyle y' = \boxed{2 {e}^(2x) + (1)/(x) {e}^(2x) + 2 \ln(x) {e}^(2x) }`

and we're done!