What is the derivative of #y=pi^x+x^pi#?

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Answer 1 · 2016-09-29T05:04:29

#y' = ln(pi)*x^pi + pix^(pi-1)#

#y=pi^x+x^pi#

Let's split this into two: #y = y_1=pi^x# plus #y_2 =x^pi#

#lny_1= xlnpi#

#:.y_1' = lnpi * x^pi#

#y_2' = pix^(pi-1)# By Power rule

Now: #y' = y_1' + y_2'#

Hence: #y' = ln(pi)*x^pi + pix^(pi-1)#