How do you differentiate $5^{2 x^{2}}$ $5^(2x^2)$ ?

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How do you differentiate $5^{2 x^{2}}$ $5^(2x^2)$ ?

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Answer 1 · 2017-10-30T16:41:52

$\frac{d}{d x} (5^{2 x^{2}}) = 4 ln 5 x 5^{2 x^{2}}$ $d/dx (5^(2x^2)) = 4ln5 x 5^(2x^2)$

Explanation:

Consider that:

$5^{2 x^{2}} = {(e^{ln 5})}^{2 x^{2}} = e^{2 ln 5 x^{2}}$ $5^(2x^2) = (e^ln5)^(2x^2) = e^(2ln5x^2)$

So, using the chain rule:

$\frac{d}{d x} (5^{2 x^{2}}) = \frac{d}{d x} (e^{2 ln 5 x^{2}}) = e^{2 ln 5 x^{2}} \frac{d}{d x} (2 ln 5 x^{2})$ $d/dx (5^(2x^2)) = d/dx(e^(2ln5x^2)) = e^(2ln5x^2) d/dx (2ln5x^2)$

$\frac{d}{d x} (5^{2 x^{2}}) = \frac{d}{d x} (e^{2 ln 5 x^{2}}) = 4 ln 5 x e^{2 ln 5 x^{2}} = 4 ln 5 x 5^{2 x^{2}}$ $d/dx (5^(2x^2)) = d/dx(e^(2ln5x^2)) = 4ln5xe^(2ln5x^2) = 4ln5 x 5^(2x^2)$

Answer 2 · 2017-10-31T08:35:30

$\frac{d y}{d x} = 5^{2 x^{2}} 4 x ln 5$ $(dy)/(dx)=5^(2x^2)4xln5$

Explanation:

$y = 5^{2 x^{2}}$ $y=5^(2x^2)$

take natural logs of both sides

$ln y = {ln 5}^{2 x^{2}}$ $lny=ln5^(2x^2)$

using laws of logs

$ln y = 2 x^{2} ln 5$ $lny=2x^2ln5$

differentiate implicitly

$\frac{1}{y} \frac{d y}{d x} = 4 x ln 5$ $1/y(dy)/(dx)=4xln5$

$\frac{d y}{d x} = y 4 x ln 5$ $(dy)/(dx)=y4xln5$

$:.(dy)/(dx)=5^(2x^2)4xln5$

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