What is the derivative of cos x raised to the x?

Answer 1

cos(xx)?

d/dx(cosu)=-sin(u)*d/dx(u)

d/dx(cos(xx))=-sin(xx)*d/dx(xx)

-The derivative of xx is:

y=xx ;You have to use implicit derivation because there is no formula for taking the derivative of uu.

lny=lnxx

lny=xlnx

-The derivative of lnx is:

d/dx(lnu)=(1/u)*d/dx(u)

-d/dx(uv)= u*dv/dx+v*du/dx

Therefore:

(1/y)*dy/dx=x*[(1/x)*d/dx(x)]+lnx(d/dx(x))

-The derivative of x is:

d/dx(xn)=nxn-1

d/dx(x)=1*x1-1

d/dx(x)=1*x0

d/dx(x)=1*(1)

d/dx(x)=1

(1/y)*dy/dx=x*[(1/x)*(1)]+lnx(1)

(1/y)*dy/dx=x*[(1/x)]+lnx

(1/y)*dy/dx=(x/x)+lnx

(1/y)*dy/dx=1+lnx

dy/dx=y(1+lnx) ;Multiply y to both sides

dy/dx=xx(1+lnx) ;y=xx, so replace the y with xx

d/dx(cos(xx))=-sin(xx)*[xx*(1+lnx)]

d/dx(cos(xx))=-(1+lnx)*xx*sin(xx)

(cosx)x?

Again with the implicit derivation:

y=(cosx)x

lny=x*ln(cosx)

(1/y)*dy/dx=x[d/dx(lncosx)]+lncosx(d/dx(x))

(1/y)*dy/dx=x[(1/cosx)*(-sinx)(1)]+lncosx(1) ;The derivative of lncosx is (1/cosx)*d/dx(cosx). The derivative of cosx is (-sinx)*d/dx (x). The derivative of x is 1.

(1/y)*dy/dx=x[(1/cosx)*(-sinx)]+lncosx

(1/y)*dy/dx=x[-tanx]+lncosx

(1/y)*dy/dx=-xtanx+lncosx

dy/dx=y(-xtanx+lncosx) ;Multiply both sides by y

dy/dx=(cosx)x(-xtanx+lncosx) ;y=(cosx)x, replace all y's with (cosx)x

dy/dx=(cosx)x(-xtanx+lncosx)=(cosx)x-1(cosx*lncosx-xsinx)