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))#