What is the derivative of #sqrttan x #?

2 Answers
Dec 18, 2015

#f'(x)=1/(2sec^2x)#

Explanation:

Original Equation:
#f(x)=sqrt(tanx)#

Rearrange:
#tanx^(1/2)#

Use Chain Rule to derive:
#(1/2)(sec^2x)(tanx)^(-1/2)#

Rearrange to make exponents positive:
#(sec^2x)/(2(tanx)^(1/2)#

Your final answer should be:
#f'(x)=(sec^2x)/(2sqrttanx#

Dec 22, 2015

#sec^2x/(2sqrttanx#

Explanation:

First rewrite the function:

#f(x)=(tanx)^(1/2)#

According to the chain rule,

#d/dx[u^(1/2)]=1/2u^(-1/2)*u'#

so,

#d/dx[(tanx)^(1/2)]=1/2(tanx)^(-1/2)*d/dx[tanx]#

#=>1/(2sqrttanx)*sec^2x#

#=>sec^2x/(2sqrttanx#