Answer:
[tex] \rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} [/tex]
Step-by-step explanation:
we would like to figure out the differential coefficient of [tex]e^{2x}(1+\ln(x))[/tex]
remember that,
the differential coefficient of a function y is what is now called its derivative y', therefore let,
[tex] \displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )[/tex]
to do so distribute:
[tex] \displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x} [/tex]
take derivative in both sides which yields:
[tex] \displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )[/tex]
by sum derivation rule we acquire:
[tex] \rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x} [/tex]
Part-A: differentiating $e^{2x}$
[tex] \displaystyle \frac{d}{dx} {e}^{2x} [/tex]
the rule of composite function derivation is given by:
[tex] \rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)[/tex]
so let g(x) [2x] be u and transform it:
[tex] \displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x[/tex]
differentiate:
[tex] \displaystyle {e}^{u} \cdot 2[/tex]
substitute back:
[tex] \displaystyle \boxed{2{e}^{2x} }[/tex]
Part-B: differentiating ln(x)•e^2x
Product rule of differentiating is given by:
[tex] \displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)[/tex]
let
- [tex]f(x) \implies \ln(x) [/tex]
- [tex]g(x) \implies {e}^{2x} [/tex]
substitute
[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x} [/tex]
differentiate:
[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }[/tex]
Final part:
substitute what we got:
[tex] \rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }[/tex]
and we're done!
