Find the derivative of the following function h(t) = \frac{t^{\frac{1}{2}}}{t^2-3t+1}

Question:

Find the derivative of the following function {eq}h(t) = \frac{t^{\frac{1}{2}}}{t^2-3t+1} {/eq}

Answer and Explanation:

Let {eq}h(t)=\frac{t^{\frac{1}{2}}}{t^2-3t+1} {/eq}. We compute the derivative using the quotient rule recalling that {eq}\dfrac{d}{dt}[t^n]=nt^{n-1} {/eq} for all {eq}n {/eq}.

{eq}\begin{align} h'(t)&=\dfrac{d}{dt}[\frac{t^{\frac{1}{2}}}{t^2-3t+1} ]\\ &=\dfrac{\frac{d}{dt}[t^{\frac{1}{2}} ](t^2-3t+1)-t^{\frac{1}{2}}\frac{d}{dt}[t^2-3t+1]}{(t^2-3t+1)^2}\\ &=\dfrac{\frac{1}{2}t^{\frac{-1}{2}}(t^2-3t+1)-t^{\frac{1}{2}}(2t-3)}{(t^2-3t+1)^2}\\ &=\dfrac{(t^2-3t+1)-2t(2t-3)}{2t^{\frac{1}{2}}(t^2-3t+1)^2}\\ &=\dfrac{-3t^2+3t+1}{2t^{\frac{1}{2}}(t^2-3t+1)^2} \end{align} {/eq}


Learn more about this topic:

Quotient Rule: Formula & Examples

from Division: Help & Review

Chapter 1 / Lesson 5
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