Question

Use the logarithmic differentiation to find the derivative f(x)=(cos2x)^lnx

Answer 1

y=(cos2x)^lnx
lny=ln(cos2x)^lnx
lny=lnx*ln(cos2x)
y'/y=1/x*ln(cos2x)+lnx*-2sin2x/cos2x
y'=y*(1/x*ln(cos2x)+lnx*-2sin2x/cos2x)
y'=(cos2x)^lnx * (1/x*ln(cos2x)+lnx*-2sin2x/cos2x

Answer 2

what does it mean *

Answer 3

\[f(x)=(\cos2x)^{lnx}\]can be rewritten as\[d(\cos^{\ln(x)}2x)/dx\]Using the chain rule with u=cos2x and v=ln(x), we get\[\ln(x)\cos^{\ln(x-1)}(2x)[d(\cos2x)/dx]+\ln(\cos2x)\cos^{\ln(x)}2x[d(\ln x)/dx]\]Use the chain rule again where u=2x and d(cos u)/du=-sin(u), we get\[\ln(\cos2x)\cos^{\ln(x)}2x[d(\ln x)]+\ln(x)\cos^{\ln(x-1)}2x(\sin2x)-[d(2x)/dx]\]Now we factor out the constants:\[\cos^{\ln(x)}(2x)\ln(\cos 2x)[d(\ln x)/dx]-\ln(x)\sin(2x)\cos^{\ln(x-1)}(2x)(2[d (\ln x)/dx])\]Where the derivative of ln(x) is (1/x), we now get\[(1/x)\cos^{\ln(x)}(2x)\ln(\cos 2x)-2\ln(x)\sin (2x)\cos^{\ln(x-1)}(2x)[d(x)/dx]\]\[=[\cos^{\ln x}(2x)\ln(\cos 2x)/x]-21\ln(x)\sin(2x)\cos^{\ln(x-1)}2x\]

I know it seems like a lot but just go slowly through it and break it down. Hopefully it'll help.