Find f'(x) if f(x) = ln [x^8 (x+9)^2 (x^2 + 5)^4 ]

Thank you very much for your help.

f(x)=ln(abc)=lna + lnb + lnc

f'(x)=1/a da/dx + 1/b (db/dx) + 1/c (dc/dx)

Go for it.

THANK YOU SO MUCH! =)

To find the derivative of f(x), we will use the rules of differentiation. In this case, we will need to apply the product rule and the chain rule.

The product rule states that if you have two functions u(x) and v(x), the derivative of their product is given by (u'(x)v(x) + u(x)v'(x)).

The chain rule states that if you have a composition of functions f(g(x)), the derivative is given by (f'(g(x)) * g'(x)).

Let's start by simplifying the expression inside the ln function:

f(x) = ln [x^8 (x+9)^2 (x^2 + 5)^4 ]

Using the product rule and the chain rule, we can find the derivative as follows:

f'(x) = [1/x^8 (x+9)^2 (x^2 + 5)^4 ] * [8x^7 (x+9)^2 (x^2 + 5)^4 ]
+ [x^8 (x+9)^2 (x^2 + 5)^4 ] * [2(x+9)(x^2 + 5)^4 ]
+ [x^8 (x+9)^2 (x^2 + 5)^4 ] * [4(x^2 + 5)^3 (2x)]

Simplifying further, we get:

f'(x) = 8x^7 (x+9)^2 (x^2 + 5)^4 / x^8 (x+9)^2 (x^2 + 5)^4
+ 2(x+9)(x^2 + 5)^4 / x^8 (x+9)^2 (x^2 + 5)^4
+ 4(x^2 + 5)^3 (2x) / x^8 (x+9)^2 (x^2 + 5)^4

Now, we can cancel out common factors:

f'(x) = 8x^7 (x+9)^2 (x^2 + 5)^4 / x^8 (x+9)^2 (x^2 + 5)^4
+ 2(x+9)(x^2 + 5)^4 / x^8 (x+9)^2 (x^2 + 5)^4
+ 4(x^2 + 5)^3 (2x) / x^8 (x+9)^2 (x^2 + 5)^4

After canceling out common factors, the final simplified answer is:

f'(x) = 8/x + 2/(x+9) + 8(x^2 + 5) / (x^3 (x+9))

Therefore, the derivative of f(x) is f'(x) = 8/x + 2/(x+9) + 8(x^2 + 5) / (x^3 (x+9)).

I hope this explanation was clear and helpful. Let me know if you have any further questions!