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Sagot :
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!
Answer:
Product Rule for Differentiation
[tex]\textsf{If }y=uv[/tex]
[tex]\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}[/tex]
Given equation:
[tex]y=e^{2x}(1+\ln x)[/tex]
Define the variables:
[tex]\textsf{Let }u=e^{2x} \implies \dfrac{du}{dx}=2e^{2x}[/tex]
[tex]\textsf{Let }v=1+\ln x \implies \dfrac{dv}{dx}=\dfrac{1}{x}[/tex]
Therefore:
[tex]\begin{aligned}\dfrac{dy}{dx} & =u\dfrac{dv}{dx}+v\dfrac{du}{dx}\\\\\implies \dfrac{dy}{dx} & =e^{2x} \cdot \dfrac{1}{x}+(1+\ln x) \cdot 2e^{2x}\\\\& = \dfrac{e^{2x}}{x}+2e^{2x}(1+\ln x)\\\\ & = \dfrac{e^{2x}}{x}+2e^{2x}+2e^{2x} \ln x\\\\& = e^{2x}\left(\dfrac{1}{x}+2+2 \ln x \right)\end{aligned}[/tex]
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