Answer :
To find the value of \( g'(5) \), we start by understanding the function \( g(x) \) and then compute its derivative.
Given:
[tex]\[ g(x) = f(5x^2 - 4x - 61) \][/tex]
To find \( g'(x) \), we need to apply the chain rule. The chain rule for derivatives tells us:
[tex]\[ g'(x) = f'(u) \cdot \frac{du}{dx} \][/tex]
where \( u = 5x^2 - 4x - 61 \).
First, let's differentiate \( u \) with respect to \( x \):
[tex]\[ u = 5x^2 - 4x - 61 \][/tex]
[tex]\[ \frac{du}{dx} = 10x - 4 \][/tex]
Next, we plug this into the chain rule formula:
[tex]\[ g'(x) = f'(5x^2 - 4x - 61) \cdot (10x - 4) \][/tex]
We now need to evaluate this expression at \( x = 5 \):
[tex]\[ g'(5) = f'(5(5^2) - 4(5) - 61) \cdot (10(5) - 4) \][/tex]
First, compute the argument of \( f' \) (i.e., the inner function):
[tex]\[ 5(5^2) - 4(5) - 61 = 5(25) - 20 - 61 = 125 - 20 - 61 = 44 \][/tex]
We now look up \( f'(44) \) in the given table:
[tex]\[ f'(44) = -5 \][/tex]
Using this value in our derivative expression:
[tex]\[ g'(5) = f'(44) \cdot (10(5) - 4) = -5 \cdot (50 - 4) = -5 \cdot 46 = -230 \][/tex]
Therefore:
[tex]\[ g'(5) = -230 \][/tex]
Given:
[tex]\[ g(x) = f(5x^2 - 4x - 61) \][/tex]
To find \( g'(x) \), we need to apply the chain rule. The chain rule for derivatives tells us:
[tex]\[ g'(x) = f'(u) \cdot \frac{du}{dx} \][/tex]
where \( u = 5x^2 - 4x - 61 \).
First, let's differentiate \( u \) with respect to \( x \):
[tex]\[ u = 5x^2 - 4x - 61 \][/tex]
[tex]\[ \frac{du}{dx} = 10x - 4 \][/tex]
Next, we plug this into the chain rule formula:
[tex]\[ g'(x) = f'(5x^2 - 4x - 61) \cdot (10x - 4) \][/tex]
We now need to evaluate this expression at \( x = 5 \):
[tex]\[ g'(5) = f'(5(5^2) - 4(5) - 61) \cdot (10(5) - 4) \][/tex]
First, compute the argument of \( f' \) (i.e., the inner function):
[tex]\[ 5(5^2) - 4(5) - 61 = 5(25) - 20 - 61 = 125 - 20 - 61 = 44 \][/tex]
We now look up \( f'(44) \) in the given table:
[tex]\[ f'(44) = -5 \][/tex]
Using this value in our derivative expression:
[tex]\[ g'(5) = f'(44) \cdot (10(5) - 4) = -5 \cdot (50 - 4) = -5 \cdot 46 = -230 \][/tex]
Therefore:
[tex]\[ g'(5) = -230 \][/tex]