What is the derivative of ${tan}^{10} (5 x)$ $tan^10( 5x)$ ?

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Answer 1 · 2017-04-27T19:49:56

$\frac{d y}{d x} = 50 {tan}^{9} 5 x {sec}^{2} 5 x$ $(dy)/(dx)=50tan^9 5xsec^2 5x$

This needs the use of the chain rule twice

$y = {tan}^{10} (5 x)$ $y=tan^10(5x)$

$let u = 5 x - - - - (1)$ $"let " u=5x----(1)$

then

$y = {tan}^{10} u$ $y=tan^10u$

$let v = tan u \Rightarrow y = v^{10}$ $"let " v=tanu=>y=v^10$

$\frac{d y}{d u} = \frac{d y}{d v} \times \frac{d v}{d u}$ $(dy)/(du)=(dy)/(dv)xx(dv)/(du)$

$\frac{d y}{d v} = 10 v^{9}$ $(dy)/(dv)=10v^9$

$\frac{d v}{d u} = {sec}^{2} u$ $(dv)/(du)=sec^2u$

$:.(dy)/(du)=10v^9xxsec^2u$

$:.(dy)/(du)=10tan^9usec^2u$

we want $(dy)/(dx)$

$(dy)/(dx)=(dy)/(du)xx(du)/(dx)$

from $(1)$

$(du)/(dx)=5$

substituting back in and substituting the $u" function back"$

$(dy)/(dx)=5xx10tan^9 5xsec^2 5x$

$(dy)/(dx)=50tan^9 5xsec^2 5x$

What is the derivative of tan^10( 5x)tan10(5x)tan^10( 5x)?