#### To determine

**To find:**

The derivative of (cosx)x with the use of logarithmic function.

#### Answer

(cosx)x[−xtanx+logcosx]

#### Explanation

**Given:**

The function is (cosx)x.

**Formulae used:**

logxx=xlogxddxu×v=udvdx+vdudxddxlogx=1xddxcosx=−sinx

Consider y=(cosx)x

Now take the log on both the sides:

logy=log(cosx)x (I)

It is known that logxx=xlogx, use it in equation number (I):

logy=x log cosx

Now differentiate both the sides with respect to *x*:

1ydydx=ddx[xlog(cosx)] (II)

It is known that ddxu×v=udvdx+vdudx, hence it will be use in equation (II):

1ydydx=xddxlog(cosx)+logcosxdxdx

It is known that ddxlogx=1x, hence use in the above equation:

1ydydx=1cosxddxcosx+log(cosx)×1

And ddxcosx=−sinx, so the equation will be:

1ydydx=−x×1cosxsinx+log(cosx)

Since, it is known that tanx=sinxcosx.

dydx=y[−xtanx+logcosx]

Substitute the value of *y* in the above equation:

dydx=(cosx)x[−xtanx+logcosx].

**Conclusion:**

Hence the derivative of (cosx)x is (cosx)x[−xtanx+logcosx].