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How do you find the derivative of #y=sqrt(3^x-2^x)#?

1 Answer

A. S. Adikesavan

Feb 19, 2017

#=1/(2y)(3^xln3-2^xln2 )#

Explanation:

Use #a^x=e^(lna x)#

#y=(e^(ln3 x)-e^(ln2 x))^(1/2)#

#y'=1/2(e^(ln3 x)-e^(ln2 x))^(-1/2)(e^(ln3 x)-e^(ln2 x))'#

#=1/2(e^(ln3 x)-e^(ln2 x))^(-1/2)(ln3e^(ln3 x)-ln2e^(ln2 x))#

#=1/(2y)(3^xln3-2^xln2 )#

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