Question

find the derivative of the function.
f(t) = (9t^2 - 2t + 3)^(3/2)
f(t) =

Explanation:

Step1: Apply chain - rule

Let $u = 9t^{2}-2t + 3$, then $y = u^{\frac{3}{2}}$. The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$.

Step2: Find $\frac{dy}{du}$

Differentiate $y = u^{\frac{3}{2}}$ with respect to $u$. Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dy}{du}=\frac{3}{2}u^{\frac{3}{2}-1}=\frac{3}{2}u^{\frac{1}{2}}$.

Step3: Find $\frac{du}{dt}$

Differentiate $u = 9t^{2}-2t + 3$ with respect to $t$. Using the power rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we get $\frac{du}{dt}=18t-2$.

Step4: Calculate $\frac{dy}{dt}$

Substitute $\frac{dy}{du}$ and $\frac{du}{dt}$ into the chain - rule formula: $\frac{dy}{dt}=\frac{3}{2}u^{\frac{1}{2}}\cdot(18t - 2)$. Replace $u$ with $9t^{2}-2t + 3$, so $f^{\prime}(t)=\frac{3}{2}(9t^{2}-2t + 3)^{\frac{1}{2}}\cdot(18t - 2)=(27t - 3)(9t^{2}-2t + 3)^{\frac{1}{2}}$.

Answer:

$(27t - 3)(9t^{2}-2t + 3)^{\frac{1}{2}}$