How do you differentiate #y= x^(11cosx)#?

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Answer 1 · 2018-05-30T15:46:35

#dy/dx=11x^(11cos(x))(cos(x)/x-sin(x)ln(x))#

#y=x^(11cos(x))#

To deal with tricky exponents like this, let's take the natural logarithm of both sides and remember the rule #log(a^b)=blog(a)#.

#ln(y)=ln(x^(11cos(x)))#

#ln(y)=11cos(x)ln(x)#

Now take the derivative on both sides. On the left, we'll need the chain rule. On the right, we'll use the product rule.

#1/y(dy/dx)=11(d/dxcos(x))ln(x)+11cos(x)(d/dxln(x))#

#1/y(dy/dx)=11(-sin(x))ln(x)+11cos(x)(1/x)#

Solving for the derivative:

#dy/dx=y((11cos(x))/x-11sin(x)ln(x))#

#dy/dx=11x^(11cos(x))(cos(x)/x-sin(x)ln(x))#