# Differentiate: h(t) = {sin t^2}/{t^2}

## Question:

Differentiate: {eq}h(t) = \dfrac {\sin t^2}{t^2} {/eq}

## Rules to be used for differentiation:

To differentiate the given function we are going to use the chain rule, quotient rule and the power rule.

Chain rule states the derivative of {eq}m(n(x)) {/eq} is {eq}m'(n(x)) \cdot n'(x) {/eq}.

Quotient rule states that the derivative of {eq}\left ( \dfrac uv \right ) {/eq} is {eq}\left [ \dfrac {u'v - v'u}{v^2} \right ] {/eq}.

Power rule states that the derivative of {eq}x^n {/eq} is {eq}nx^{n - 1} {/eq}, where {eq}n {/eq} is a real number.

We have,

$$h(t) = \dfrac {\sin t^2}{t^2}$$

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

\displaystyle \begin{align*} h'(t) &= \dfrac {d}{dt} \left [ \frac{\sin t^2}{t^2} \right ] \\ &= \dfrac {\dfrac {d}{dt} \left ( \sin t^2 \right ) \cdot t^2 - \dfrac {d}{dt} \left ( t^2 \right ) \cdot \sin t^2}{\left ( t^2 \right )^2} \\ &= \dfrac {\cos t^2 \cdot \dfrac {d}{dt} \left ( t^2 \right ) \cdot t^2 - \left ( 2t \right ) \cdot \sin t^2}{t^4} \\ &= \dfrac {\cos t^2 \cdot (2t) \cdot t^2 - 2t \ \sin t^2}{t^4} \\ &= \dfrac {2t^3 \ \cos t^2 - 2t \ \sin t^2}{t^4} \\ h'(t) &= \dfrac {2t^2 \ \cos t^2 - 2 \ \sin t^2}{t^3} \\ \end{align*}