How do you find the derivative of #x^(log(base5)(x))#?

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Answer 1 · 2016-09-18T14:14:19

#2x^(log_5(x)-1)log_5(x)#

#y=x^(log_5(x))#

Take the natural logarithm of both sides. (This is known as logarithmic differentiation.)

#ln(y)=ln(x^(log_5(x)))#

Use the rule: #log(a^b)=b*log(a)# to rewrite this function:

#ln(y)=log_5(x)*ln(x)#

Now, rewrite #log_5(x)# in logarithms with base #e# using the change of base formula: #log_a(b)=log_c(b)/log_c(a)#.

#ln(y)=ln(x)/ln(5)*ln(x)#

#ln(y)=(ln(x))^2/ln(5)#

Differentiate both sides. The chain rule will be needed on both sides of the equation.

#1/y*dy/dx=(2ln(x))/ln(5)*d/dxln(x)#

We already have used the derivative of #ln(x)# is #1/x#. Also, rewrite #y# as #x^(log_5(x))#:

#1/x^(log_5(x))*dy/dx=(2ln(x))/(xln(5))#

Note that #ln(x)/ln(5)=log_5(x)# through the change of base formula in reverse, or #log_c(b)/log_c(a)=log_a(b)#.

#1/x^(log_5(x))*dy/dx=(2log_5(x))/x#

Multiplying both sides by #x^(log_5(x))#:

#dy/dx=(2x^(log_5(x))log_5(x))/x#

Note that #x^(log_5(x))/x=x^(log_5(x)-1)# through the rule that #x^a/x^b=x^(a-b)#:

#dy/dx=2x^(log_5(x)-1)log_5(x)#