Question

Find dy/dx if y=e^2x.log(x+1)

Answer 1

Answer:

The value of the given differential is $\frac{dy}{dx} =e^{2x}(\frac{1+2(x+1)log(x+1)}{x+1} )$

Step-by-step explanation:

The given function is $y=e^{2x}log(x+1)$

To differentiate the function,we shall use the product rule as shown below

$\frac{d}{dx} (ab)=a\frac{d}{dx}(b)+b\frac{d}{dx}(a)$

Therefore,the differentiation of the given function becomes,

$\frac{dy}{dx} =e^{2x}\frac{d}{dx}[log(x+1)]+log(x+1)\frac{d}{dx}(e^{2x})$

We have basic relations for differentiation as mentioned below

$\frac{d}{dx}(e^{x})=e^{x}\\\frac{d}{dx}(logx)=\frac{1}{x}$

Using these relation in our differentiation,we get

$\frac{dy}{dx} =e^{2x}(\frac{1}{x+1}) +log(x+1)(2e^{2x})\\=e^{2x}(\frac{1+2(x+1)log(x+1)}{x+1} )$
Therefore,the value of the given differential is $\frac{dy}{dx} =e^{2x}(\frac{1+2(x+1)log(x+1)}{x+1} )$

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Answer 2

Answer:

The derivative dy/dx of the function $y=e^{2x}log(x+1)$ is $e^{2x}\left(\frac{1}{x+1}+2 log(x+1)\right)$.

Step-by-step explanation:

Recall the derivatives of exponential function and logarithm function,

1) The derivative of the exponential function is,

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

2) The derivative of the logarithm function is,

$\frac{d}{dx}(logx)=\frac{1}{x}$

Step 1 of 2

Consider the given function as follows:

$y=e^{2x}log(x+1)$

Differentiate both the sides with respect to $x$ as follows:

$\frac{dy}{dx} =\frac{d}{dx}\left[e^{2x}log(x+1)\right]$

Using the product rule, differentiate the right-hand side as follows:

$\frac{dy}{dx}=e^{2x}\frac{d}{dx}(log(x+1))+log(x+1)\frac{d}{dx}(e^{2x})$

$\frac{dy}{dx}=e^{2x}\times\frac{1}{x+1}+log(x+1)\times2e^{2x}$

Step 2 of 2

Find the value of dy/dx.

Further, simplify the right-hand side as follows:

$\frac{dy}{dx}=\frac{e^{2x}}{x+1}+2e^{2x}\cdot log(x+1)$

Take the exponential function $e^{2x}$ common out, we get

$\frac{dy}{dx}=e^{2x}\left(\frac{1}{x+1}+2 log(x+1)\right)$

Final answer: The derivative dy/dx of the function $y=e^{2x}log(x+1)$ is $e^{2x}\left(\frac{1}{x+1}+2 log(x+1)\right)$.

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