How do you take the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?

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How do you take the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?

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Answer 1 · 2018-06-19T13:23:48

$\frac{d y}{d x} = y {\frac{4 tan (5 x)}{sin (4 x) cos (4 x)} + 5 {sec}^{2} (5 x) ln tan (4 x)}$ $(dy)/(dx)=y{(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)}$

Explanation:

Let,

$y = {tan (4 x)}^{tan (5 x)}$ $y=tan(4x)^tan(5x)$

Taking natural log. ,both sides

$ln y = {ln tan (4 x)}^{tan (5 x)}$ $lny=lntan(4x)^tan(5x)$

$ln y = tan (5 x) \cdot ln tan (4 x)$ $lny=tan(5x)*lntan(4x)$

Diff.w.r.t $x$ $x$ ,Using Product Rule:

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

$\frac{1}{y} \frac{d y}{d x} = tan (5 x) \cdot \frac{4}{tan (4 x)} {sec}^{2} (4 x) + ln tan (4 x) 5 {sec}^{2} (5 x)$ $1/y(dy)/(dx)=tan(5x)*4/tan(4x)sec^2(4x)+lntan(4x)5sec^2(5x)$

$\frac{1}{y} \frac{d y}{d x} = 4 tan (5 x) \frac{cos (4 x)}{sin (4 x)} \times \frac{1}{{cos}^{2} (4 x)} + 5 {sec}^{2} (5 x) ln t a n (4 x)$ $1/y(dy)/(dx)=4tan(5x)cos(4x)/sin(4x)xx1/cos^2(4x)+5sec^2(5x)lnt an(4x)$

$\frac{1}{y} \frac{d y}{d x} = \frac{4 tan (5 x)}{sin (4 x) cos (4 x)} + 5 {sec}^{2} (5 x) ln tan (4 x)$ $1/y(dy)/(dx)=(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)$

$\frac{d y}{d x} = y {\frac{4 tan (5 x)}{sin (4 x) cos (4 x)} + 5 {sec}^{2} (5 x) ln tan (4 x)}$ $(dy)/(dx)=y{(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)}$

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How do you take the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?

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Explanation:

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Humanities

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How do you take the derivative of tan(4x)^tan(5x)tan(4x)tan(5x)tan(4x)^tan(5x)?

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How do you take the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?