Questions: Find the derivative of the function. f(t)=4 sqrt(2 t^2+7) f'(t)=

Solution Steps

To find the derivative of the function \( f(t) = 4 \sqrt{2t^2 + 7} \), we will use the chain rule. The chain rule states that if you have a composite function \( f(g(t)) \), then the derivative is \( f'(g(t)) \cdot g'(t) \). Here, we can identify the outer function as \( 4 \sqrt{u} \) and the inner function as \( u = 2t^2 + 7 \).

We are given the function: \[ f(t) = 4 \sqrt{2t^2 + 7} \]

To find the derivative \( f'(t) \), we use the chain rule. Let: \[ u = 2t^2 + 7 \] Then: \[ f(t) = 4 \sqrt{u} \]

The derivative of the outer function \( 4 \sqrt{u} \) with respect to \( u \) is: \[ \frac{d}{du} \left( 4 \sqrt{u} \right) = 4 \cdot \frac{1}{2} u^{-\frac{1}{2}} = \frac{2}{\sqrt{u}} \]

The derivative of the inner function \( 2t^2 + 7 \) with respect to \( t \) is: \[ \frac{d}{dt} \left( 2t^2 + 7 \right) = 4t \]

Using the chain rule, we multiply the derivatives of the outer and inner functions: \[ f'(t) = \frac{2}{\sqrt{2t^2 + 7}} \cdot 4t = \frac{8t}{\sqrt{2t^2 + 7}} \]

Final Answer