How do you find the derivative of $sqrt(xtanx)$ ?

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Answer 1 · 2017-02-05T14:16:15

$d/dx sqrt(xtanx) = (x +tanx +xtan^2x)/(2sqrt(xtanx))$

$d/dx f(y(x)) = (df)/dy * dy/dx$

So, given:

$f(x) = sqrt(xtanx)$

we pose $y=x tanx$ , so $f(y) = sqrt(y)$ and $(df)/dy = 1/(2sqrty)$ and we have:

$d/dx sqrt(xtanx) = 1/(2sqrt(xtanx)) d/dx (xtanx)$

Now we calculate this derivative using the product rule:

$d/dx (xtanx) = (d/dx x) tanx + x (d/dx tanx)$

$d/dx (xtanx) = tanx + x / cos^2x$

we try to simplify this expression using the trigonometric identity:

$1/cos^2x = 1+tan^2x$

$d/dx (xtanx) = x +tanx +xtan^2x$

Putting it together:

$d/dx sqrt(xtanx) = (x +tanx +xtan^2x)/(2sqrt(xtanx))$

How do you find the derivative of sqrt(xtanx)sqrt(xtanx)?