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Find f'(3), if f(z) = z^2 + 1/z.

Question:

Find {eq}f'(3), {/eq} if {eq}f(z) = \dfrac {z^2 + 1}{z}. {/eq}

Quotient rule of differentiation:


Let us consider {eq}r {/eq} and {eq}s {/eq} are differentiable, then {eq}g(t) = \dfrac {r(t)}{s(t)} {/eq} is also differentiable.

Then the derivative of {eq}g(t) {/eq} is given by,

{eq}g'(t) = \dfrac {r'(t) \cdot s(t) - s'(t) \cdot r(t)}{(s(t))^2} {/eq}

Answer and Explanation:


{eq}\displaystyle f(z) = \frac{z^2 + 1}{z} {/eq}


Differentiating the function with respect to {eq}z {/eq},

$$\displaystyle \begin{align*} f'(z) &= \frac {d}{dz} (f(z)) \\ &= \dfrac {d}{dz} \left ( \dfrac {z^2 + 1}{z} \right ) \\ &= \dfrac {\dfrac {d}{dz} \left ( z^2 + 1 \right ) \cdot z - \dfrac {d}{dz} (z) \cdot \left ( z^2 + 1 \right )}{(z)^2} \\ &= \dfrac {\left ( 2z + 0 \right ) \cdot z - (1) \cdot \left ( z^2 + 1 \right )}{z^2} \\ &= \dfrac {\left ( 2z \right ) z - \left ( z^2 + 1 \right )}{z^2} \\ &= \dfrac {2z^2 - z^2 - 1 }{z^2} \\ f'(z) &= \dfrac { z^2 - 1 }{z^2} \\ \end{align*} $$


At {eq}z = 3 {/eq}, we get,

$$\displaystyle \begin{align*} f'(3) &= \dfrac { (3)^2 - 1 }{(3)^2} \\ &= \dfrac { 9 - 1 }{9} \\ f'(3) &= \dfrac 89 \\ \end{align*} $$


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Quotient Rule: Formula & Examples

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Chapter 1 / Lesson 5
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