Question #e8ba4

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Question #e8ba4

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Answer 1 · 2017-01-24T17:59:28

$\frac{d}{d x} {(sin x)}^{tan x} = {(sin x)}^{tan x} (\frac{ln (sin x)}{{cos}^{2} x} + 1)$ $d/(dx) (sinx)^tanx = (sinx)^tanx (ln (sinx)/cos^2x +1)$

Explanation:

In cases like this the easiest approach is to pass through logarithms:

${(sin x)}^{tan x} = {(e^{ln (sin x)})}^{tan x} = e^{tan x \cdot ln (sin x)}$ $(sinx)^tanx = (e^ln (sinx) )^tanx = e^(tanx*ln(sinx))$

so:

$\frac{d}{d x} {(sin x)}^{tan x} = e^{tan x \cdot ln (sin x)} \cdot \frac{d}{d x} (tan x \cdot ln (sin x))$ $d/(dx) (sinx)^tanx = e^(tanx*ln(sinx))* d/dx(tanx*ln(sinx))$

$\frac{d}{d x} {(sin x)}^{tan x} = {(sin x)}^{tan x} (\frac{ln (sin x)}{{cos}^{2} x} + tan x \frac{cos x}{sin x})$ $d/(dx) (sinx)^tanx = (sinx)^tanx (ln (sinx)/cos^2x + tanx cosx/sinx)$

$\frac{d}{d x} {(sin x)}^{tan x} = {(sin x)}^{tan x} (\frac{ln (sin x)}{{cos}^{2} x} + 1)$ $d/(dx) (sinx)^tanx = (sinx)^tanx (ln (sinx)/cos^2x +1)$

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Question #e8ba4

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