How do you find the derivatives of $y = {(2 x + 1)}^{e}$ $y=(2x+1)^e$ by logarithmic differentiation?

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How do you find the derivatives of $y = {(2 x + 1)}^{e}$ $y=(2x+1)^e$ by logarithmic differentiation?

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Answer 1 · 2017-03-08T14:35:36

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

Explanation:

Take the natural log of both sides.

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

$ln y = e ln (2 x + 1)$ $lny = eln(2x + 1)$

Now use implicit differentiation and the product rule.

$\frac{1}{y} (\frac{d y}{d x}) = 0 (ln (2 x + 1)) + \frac{2 e}{2 x + 1}$ $1/y(dy/dx) = 0(ln(2x + 1)) + (2e)/(2x + 1)$

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

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

By the quotient rule of exponents:

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

Hopefully this helps!

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How do you find the derivatives of $y = {(2 x + 1)}^{e}$ $y=(2x+1)^e$ by logarithmic differentiation?

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Explanation:

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How do you find the derivatives of $y = {(2 x + 1)}^{e}$ $y=(2x+1)^e$ by logarithmic differentiation?