How to determine dy/dx if #y=(cosx)^π#?

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Answer 1 · 2018-05-23T14:20:55

# dy/dx=-pi*sinx*(cosx)^(pi-1)#.

#y=(cosx)^pi#.

Using the Power Rule and the Chain Rule, we get,

#dy/dx=pi(cosx)^(pi-1)*d/dx{cosx}#.

#:. dy/dx=-pi*sinx*(cosx)^(pi-1)#.

Answer 2 · 2018-05-23T14:24:18

#(dy)/(dx)=-pisinx(cosx)^(pi-1)#

We can use here concept of function of a function.

Let #f(x)=x^pi# and #g(x)=cosx#

then #(df)/(dx)=pix^(pi-1)# and #(dg)/(dx)=-sinx#

Now as #y=f(g(x))=(g(x))^pi# we have #(dy)/(dg(x))=pi(g(x))^(pi-1)#

and as #(dy)/(dx)=(dy)/(dg(x))*(dg(x))/dx)#

= #pi(g(x))^(pi-1)*(-sinx)#

= #-pisinx(cosx)^(pi-1)#