Questions: If h(x)=u(x)^v(x), then d/dx h(x)=u(x)^v(x)[v(x) u'(x)/u(x)+(ln(u(x)) v'(x))]. Use the above definition to find the derivative of the following function. h(x)=(x^2+4)^(2x) d/dx h(x)=□

Solution Steps

To find the derivative of the function $h(x) = (x^2 + 4)^{2x}$ using the given formula, we need to identify $u(x)$ and $v(x)$ and then apply the formula. Here, $u(x) = x^2 + 4$ and $v(x) = 2x$. We will then compute $u'(x)$ and $v'(x)$, and substitute these into the formula to find the derivative.

1. Identify $u(x)$ and $v(x)$.
2. Compute $u'(x)$ and $v'(x)$.
3. Substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the given derivative formula.
4. Simplify the expression to get the final derivative.

Let $u(x) = x^2 + 4$ and $v(x) = 2x$.

The derivatives are calculated as follows:

• $u'(x) = \frac{d}{dx}(x^2 + 4) = 2x$
• $v'(x) = \frac{d}{dx}(2x) = 2$

Using the formula for the derivative of $h(x) = u(x)^{v(x)}$: $$\frac{d}{dx} h(x) = u(x)^{v(x)} \left[ \frac{v(x) u'(x)}{u(x)} + \ln(u(x)) v'(x) \right]$$ Substituting the values: $$\frac{d}{dx} h(x) = (x^2 + 4)^{2x} \left[ \frac{2x \cdot 2x}{x^2 + 4} + \ln(x^2 + 4) \cdot 2 \right]$$

The expression simplifies to: $$\frac{d}{dx} h(x) = (x^2 + 4)^{2x} \left[ \frac{4x^2}{x^2 + 4} + 2 \ln(x^2 + 4) \right]$$ Further simplification yields: $$\frac{d}{dx} h(x) = 2(x^2 + 4)^{2x - 1} \left[ 2x^2 + (x^2 + 4) \ln(x^2 + 4) \right]$$

Final Answer