How do you differentiate ln[(x(^2)+5)^4]?

Question

Im really not sure how to do it. Could you please show steps. Thanks :)

Answer 1

the basic rule here is that d/dt lnt = 1/t .. in this case t= [(x^2+5)^4] so...

d/dx ln (x^2+5)^4 = [1/(x^2+5)^4] * [4(x^2+5)*2x] (the last bit because of the chain rule)

Answer 2

f(x) = ln[(x² + 5)^4]

Remember, using log laws, you can bring out that 4:

f(x) = 4*ln(x² + 5)

Product rule:

u = 4

u' = 0

v = ln(x² + 5)

v' = 2x / (x² + 5)

f'(x) = u'v + uv'

f'(x) = 4*[2x / (x² + 5)]

f'(x) = 8x / (x² + 5)

In fact, we didn't need to use the product rule but it's good to show you what I did with the 4.

Remember:

The differentiation of ln f(x) is f'(x) / f(x)

Answer 3

Well this is going to be quite involved.

First of all, the derivative of lnx=1/x We'll treat (x^2+5)^4 as x.

So we get 1/(x^2+5)^4

Then we have to use the Chain Rule and differentiate (x^2+5)^4 and multiply.

The derivative of (x^2+5)^4 is:

(4(x^2+5)^3)*2x

So we multiply:

(1/(x^2+5)^4)*((4(x^2+5)^3)*2x)

And we get:

(8x)/(x^2+5)

Any questions, let me know.

Answer 4

I don't know