Use logarithmic differentiation to find the derivative of the function. y = x^8cosx

y = x^ 8cosx

y'=?

I am assuming that the function is:
y = x^[8cos(x)].

Start out by taking the natural logarithm of both sides to get:
ln(y) = ln{x^[8cos(x)]}
==> ln(y) = 8cos(x)*ln(x), since ln(a^b) = b*ln(a).

Then, by differentiating implicitly:
y'/y = -8sin(x)*ln(x) + 8cos(x)/x, by applying the Product Rule to the right side
==> y'/y = [8cos(x) - 8x*sin(x)*ln(x)]/x, by combining fractions
==> y' = y * [8cos(x) - 8x*sin(x)*ln(x)]/x, by multiplying both sides by y
==> y' = x^[8cos(x)] * [8cos(x) - 8x*sin(x)*ln(x)]/x, since y = x^[8cos(x)]
==> y' = x^[8cos(x) - 1] * [8cos(x) - 8x*sin(x)*ln(x)], by simplifying.

I hope this helps!

y = x^(8∙cos(x))

ln(y) = 8∙cos(x)∙ln(x)

1/y * y' = 8 * [cos(x)*1/x - sin(x)∙ln(x)]
y' = 8y * [cos(x)*1/x - sin(x)∙ln(x)]

y' = 8x^(8∙cos(x)) * [cos(x)*1/x - sin(x)∙ln(x)]

If you were to pay attention in school, I'd bet you could answer this.