Question

differentiate the function. (h(x)=(x - 2)(2x + 3)) (h(x)=) 7. -/1 points differentiate the function. (g(t)=4t^{-3/8}) (g(t)=)

Explanation:

Step1: Expand the function h(x)

\[

$$\begin{align*} h(x)&=(x - 2)(2x+3)\\ &=x(2x + 3)-2(2x + 3)\\ &=2x^{2}+3x-4x - 6\\ &=2x^{2}-x - 6 \end{align*}$$

\]

Step2: Differentiate h(x) using power - rule

The power - rule states that if \(y = ax^{n}\), then \(y^\prime=anx^{n - 1}\).
For \(h(x)=2x^{2}-x - 6\), \(h^\prime(x)=2\times2x^{2 - 1}-1\times x^{1 - 1}-0 = 4x-1\)

Step3: Differentiate g(t) using power - rule

For \(g(t)=4t^{-\frac{3}{8}}\), by the power - rule \(g^\prime(t)=4\times(-\frac{3}{8})t^{-\frac{3}{8}-1}\)
\[

$$\begin{align*} g^\prime(t)&=4\times(-\frac{3}{8})t^{-\frac{3 + 8}{8}}\\ &=-\frac{3}{2}t^{-\frac{11}{8}} \end{align*}$$

\]

Answer:

\(h^\prime(x)=4x - 1\), \(g^\prime(t)=-\frac{3}{2}t^{-\frac{11}{8}}\)