Question

Find the derivative using logarithmic differentiation?

Answer 1

`dy/dx=4x^(4x)(lnx+1)`

Explanation:

Take the natural log of both sides.

`lny=lnx^(4x)`

`loga^b=bloga`. So,

`lny=4xlnx`

Take the derivative of both sides.

`(1/y)dy/dx=4+4lnx`

Multiply both sides by `y` and substitute in the original equation.

`dy/dx=4x^(4x)(lnx+1)`

I'm not here to judge, but I noticed this is the second logarithmic differentiation problem I have answered for you. Perhaps, you should try to better understand logarithmic differentiation. It is actually quite simple.

Here is a helpful video if you are interested in learning more.

Answer 2

`d/dx x^(4x) = 4x^(4x)(1+lnx)`

Explanation:

Pose:

`f(x) = x^(4x)`

`g(x) = ln f(x) = 4xlnx`

Then:

`g'(x) = 4+4lnx = (f'(x))/f(x)`

and:

`f'(x) = f(x)g'(x) = 4x^(4x)(1+lnx)`

Answer 3

Given: `y = x^(4x)`

Use the natural logarithm on both sides:

`ln(y) = ln(x^(4x))`

Use the property of logarithms `ln(x^(4x)) = (4x)ln(x)`:

`ln(y) = (4x)ln(x)`

Differentiate both sides:

`(d(ln(y)))/dx = (d((4x)ln(x)))/dx`

The left side requires the use of the chain rule:

`1/ydy/dx = (d((4x)ln(x)))/dx`

The right side requires the use of the product rule:

`1/ydy/dx = (d(4x))/dxln(x) + 4x(d(ln(x)))/dx`

For the first term we use `(d(4x))/dx = 4`

`1/ydy/dx = 4ln(x) + 4x(d(ln(x)))/dx`

For the second term we use `(d(ln(x)))/dx = 1/x`

`1/ydy/dx = 4ln(x) + 4x1/x`

The second term simplifies:

`1/ydy/dx = 4ln(x) + 4`

Multiply both sides by y:

`dy/dx = (4ln(x) + 4)y`

Substitute `x^(4x)` for y:

`dy/dx = (4ln(x) + 4)x^(4x)`