# Differentiate y=(x+1)^(2x)?

Mar 25, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\left(x + 1\right)}^{2 x} \ln \left(x + 1\right) + 2 x {\left(x + 1\right)}^{2 x - 1}$

#### Explanation:

As $y = {\left(x + 1\right)}^{2 x}$

$\ln y = 2 x \ln \left(x + 1\right)$

and hence taking implicit differential on both sides, we get

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(x + 1\right) + 2 \frac{x}{x + 1}$

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \left(\ln \left(x + 1\right) + \frac{x}{x + 1}\right)$

$= 2 {\left(x + 1\right)}^{2 x} \left(\ln \left(x + 1\right) + \frac{x}{x + 1}\right)$

$= 2 {\left(x + 1\right)}^{2 x} \ln \left(x + 1\right) + 2 x {\left(x + 1\right)}^{2 x - 1}$

Mar 25, 2017

#### Explanation:

Given: $y = {\left(x + 1\right)}^{2 x}$

Use the natural logarithm on both sides:

$\ln | y | = \ln | {\left(x + 1\right)}^{2 x} |$

Use the identity $\ln | {b}^{a} | = \left(a\right) \ln | b |$ on the right:

$\ln | y | = \left(2 x\right) \ln | \left(x + 1\right) |$

Differentiate both sides:

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln | x + 1 | + \frac{2 x}{x + 1}$

Please observe that the left side is obtained, using implicit differentiation, and the right side is obtained, using the product rule.

We can eliminate the absolute value of we bring the 2 inside as an exponent:

1/ydy/dx= ln((x+1)^2) + (2x)/(x+1); x!=-1

Multiply both sides by y:

dy/dx= (ln((x+1)^2) + (2x)/(x+1))y; x!=-1

Substitute for y:

dy/dx= (ln((x+1)^2) + (2x)/(x+1))(x+1)^(2x); x!=-1