Use logarithmic differentiation to find dy/dx

Question

e^2y=(sinx)^(lnx)

Answer 1

e^2y = (sinx)^lnx

Taking the natural logarithm of both sides gives

2y = ln[(sinx)^lnx] = lnx × ln(sinx) (using the power rule of logarithms)

Differentiating the left-hand-side gives 2 × dy/dx

The right-hand-side = u × v where u= lnx and v = ln(sinx)

To find the derivative we use the product rule

d(u×v)dx = v×(du/dx) + u×(dv/dx)

du/dx = 1/x so v×(du/dx) = ln(sinx)/x

dv/dx = 1/sinx × cosx = cotx (using the chain rule)

so u×(dv/dx) = lnx cotx

So differentiating the right-hand-side gives

ln(sinx)/x + lnx cotx

So we have

2 × dy/dx = ln(sinx)/x + lnx cotx

Answer

dy/dx = ln(sinx)/2x + ½lnx cotx