For y(x)=ln(3+x2), find y`(6)
I keep getting 12 but it says that its the wrong answer, really stuck here
To find the derivative of the function y(x) = ln(3 + x^2), you can use the chain rule. The chain rule states that if you have a composite function, such as ln(f(x)), then the derivative is given by the derivative of the outside function (ln) multiplied by the derivative of the inside function (f'(x)).
In this case, the outside function is ln(u) and the inside function is u = 3 + x^2. To find the derivative of y(x), we need to find the derivative of ln(u) with respect to u (which is 1/u) and multiply it by the derivative of u with respect to x (which is u' = 2x).
Let's find the derivative step by step:
Step 1: Find the derivative of ln(u) with respect to u.
The derivative of ln(u) with respect to u is 1/u.
Step 2: Find the derivative of u = 3 + x^2 with respect to x.
The derivative of u = 3 + x^2 with respect to x is u' = 2x. This is because the derivative of 3 is 0, and the derivative of x^2 is 2x.
Now, we can find the derivative of y(x) by applying the chain rule:
dy/dx = (1/u) * u'
Substituting u = 3 + x^2 and u' = 2x into the equation, we have:
dy/dx = (1/(3 + x^2)) * (2x)
Now, we can find y'(6) by plugging x = 6 into the derivative dy/dx:
y'(6) = (1/(3 + 6^2)) * (2 * 6)
= (1/(3 + 36)) * 12
= (1/39) * 12
= 12/39
= 4/13
So, the value of y'(6) is 4/13, not 12.