Let [tex]\( f(x) = \ln(x^2 + 4x + 10) \)[/tex].
To find the derivative [tex]\( f'(x) \)[/tex], we will use the chain rule. The chain rule states that if you have a composite function [tex]\( h(g(x)) \)[/tex], its derivative is given by:
[tex]\[ (h(g(x)))' = h'(g(x)) \cdot g'(x) \][/tex]
In our case:
- [tex]\( h(u) = \ln(u) \)[/tex] where [tex]\( u = x^2 + 4x + 10 \)[/tex]
- [tex]\( g(x) = x^2 + 4x + 10 \)[/tex]
We will need to find the derivatives of both the outer function [tex]\( h(u) \)[/tex] and the inner function [tex]\( g(x) \)[/tex].
1. Derivative of the outer function [tex]\( h(u) = \ln(u) \)[/tex]:
The derivative of [tex]\( \ln(u) \)[/tex] with respect to [tex]\( u \)[/tex] is:
[tex]\[ h'(u) = \frac{1}{u} \][/tex]
2. Derivative of the inner function [tex]\( g(x) = x^2 + 4x + 10 \)[/tex]:
The derivative of [tex]\( x^2 + 4x + 10 \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ g'(x) = 2x + 4 \][/tex]
Now, applying the chain rule:
[tex]\[ f'(x) = h'(g(x)) \cdot g'(x) \][/tex]
Substitute [tex]\( g(x) = x^2 + 4x + 10 \)[/tex] into the derivative of [tex]\( h \)[/tex]:
[tex]\[ h'(g(x)) = \frac{1}{x^2 + 4x + 10} \][/tex]
Therefore,
[tex]\[ f'(x) = \frac{1}{x^2 + 4x + 10} \cdot (2x + 4) \][/tex]
Thus,
[tex]\[ f'(x) = \frac{2x + 4}{x^2 + 4x + 10} \][/tex]
This is the derivative of the function [tex]\( f(x) = \ln(x^2 + 4x + 10) \)[/tex]. So, the final result is:
[tex]\[ f^{\prime}(x) = \frac{2x + 4}{x^2 + 4x + 10} \][/tex]
