Question

hw 8 - derivative rules section 2.3: problem (1 point) if $f(t)=sqrt3{t^{2}} + 2sqrt{t^{3}}$, find $f(t)$. $f(t)=square$

Explanation:

Step1: Rewrite the function using exponent notation

Rewrite $\sqrt[3]{t^{2}}$ as $t^{\frac{2}{3}}$ and $2\sqrt{t^{3}}$ as $2t^{\frac{3}{2}}$. So $f(t)=t^{\frac{2}{3}} + 2t^{\frac{3}{2}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $y'=nx^{n - 1}$.
For the first term $y_1=t^{\frac{2}{3}}$, its derivative $y_1'=\frac{2}{3}t^{\frac{2}{3}-1}=\frac{2}{3}t^{-\frac{1}{3}}$.
For the second term $y_2 = 2t^{\frac{3}{2}}$, its derivative $y_2'=2\times\frac{3}{2}t^{\frac{3}{2}-1}=3t^{\frac{1}{2}}$.

Step3: Find the derivative of the whole function

Since $f(t)=y_1 + y_2$, then $f'(t)=y_1'+y_2'$.
So $f'(t)=\frac{2}{3}t^{-\frac{1}{3}}+3t^{\frac{1}{2}}$.

Answer:

$\frac{2}{3}t^{-\frac{1}{3}}+3t^{\frac{1}{2}}$