What is the derivative of (cosx)^x?

1 Answer
May 29, 2018

We use a technique called logarithmic differentiation to differentiate this kind of function.

In short, we let y = (cos(x))^x,

Then,

ln(y) = ln((cos(x))^x)

ln(y) = xln(cos(x)), by law of logarithms,

And now we differentiate.

d/dx(ln(y)) = d/dx(xln(cos(x)))

dy/dx xx d/dy(ln(y)) = ln(cos(x)) xx d/dx(x) + x d/dx(ln(cos(x)))

dy/dx xx 1/y = ln(cos(x)) - (xsin(x))/cos(x)

dy/dx = y(ln(cos(x)) - (xsin(x))/cos(x))

dy/dx = (cos(x))^x(ln(cos(x)) - (xsin(x))/cos(x))

Alternatively, you can express (cos(x))^x as e^(xln(cos(x))), but that's basically the same thing.