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Find h'(2), if h(t) = \frac{t^2 + 3}{t - 1}.

Question:

Find {eq}h'(2) {/eq}, if {eq}h(t) = \dfrac{t^2 + 3}{t - 1} {/eq}.

Quotient rule of differentiation:


Let us consider the function {eq}g(x) = \dfrac {h(x)}{i(x)} {/eq}, then the derivative of that function is given by,

{eq}g'(x) = \dfrac {h'(x) \cdot i(x) - i'(x) \cdot h(x)}{(i(x))^2} {/eq}


Power rule of differentiation:

The derivative of {eq}x^n {/eq} is {eq}nx^{n - 1} {/eq}, here {eq}n {/eq} is a real number.

Answer and Explanation:

We have,

{eq}h(t) = \dfrac{t^2 + 3}{t - 1} {/eq}

Differentiating with respect to {eq}x {/eq},

$$\displaystyle \begin{align*} h'(t) &= \dfrac {d}{dx} \left [ \dfrac{t^2 + 3}{t - 1} \right ] \\ &= \dfrac{(2t + 0) \cdot (t - 1) - (1 - 0) \cdot \left ( t^2 + 3 \right )}{(t - 1)^2} \\ &= \dfrac{2t(t - 1) - \left ( t^2 + 3 \right )}{(t - 1)^2} \\ &= \dfrac{2t^2 - 2t - t^2 - 3}{(t - 1)^2} \\ h'(t) &= \dfrac{t^2 - 2t - 3}{(t - 1)^2} \\ \Rightarrow h'(2) &= \dfrac{2^2 - 2(2) - 3}{(2 - 1)^2} \\ h'(2) &= \dfrac{4 - 4 - 3}{(1)^2} \\ h'(2) &= \dfrac{- 3}{1} \\ h'(2) &= - 3 \\ \end{align*} $$


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Quotient Rule: Formula & Examples

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Chapter 1 / Lesson 5
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