Transcript text: Given $f(x)=\frac{1}{\left(6-x^{2}\right)^{4}}$, find $f^{\prime}(x)$.
Solution
Solution Steps
To find the derivative \( f'(x) \) of the function \( f(x) = \frac{1}{(6-x^2)^4} \), we can use the chain rule and the power rule. First, rewrite the function as \( f(x) = (6-x^2)^{-4} \). Then, apply the chain rule: differentiate the outer function and multiply by the derivative of the inner function.
Step 1: Rewrite the Function
The given function is \( f(x) = \frac{1}{(6-x^2)^4} \). We can rewrite this function using a negative exponent:
\[ f(x) = (6-x^2)^{-4} \]
Step 2: Apply the Chain Rule
To find the derivative \( f'(x) \), we apply the chain rule. The chain rule states that if you have a composite function \( g(h(x)) \), then the derivative is \( g'(h(x)) \cdot h'(x) \).
Step 3: Differentiate the Outer Function
The outer function is \( g(u) = u^{-4} \), where \( u = 6-x^2 \). The derivative of \( g(u) \) with respect to \( u \) is:
\[ g'(u) = -4u^{-5} \]
Step 4: Differentiate the Inner Function
The inner function is \( h(x) = 6-x^2 \). The derivative of \( h(x) \) with respect to \( x \) is:
\[ h'(x) = -2x \]
Step 5: Combine Using the Chain Rule
Using the chain rule, the derivative \( f'(x) \) is:
\[ f'(x) = g'(h(x)) \cdot h'(x) = -4(6-x^2)^{-5} \cdot (-2x) \]
Step 6: Simplify the Expression
Simplifying the expression, we get:
\[ f'(x) = 8x(6-x^2)^{-5} \]
\[ f'(x) = \frac{8x}{(6-x^2)^5} \]