# The total worldwide box-office receipts for a long-running movie are approximated by the...

## Question:

The total worldwide box-office receipts for a long-running movie are approximated by the following function where T(x) is measured in millions of dollars and x is the number of years since the movie's release.

{eq}T(x) = \frac{120x^2}{x^2 + 4} {/eq}

How fast are the total receipts changing 1 yr, 5 yr, and 8 yr after its release? (Round your answers to two decimal places.)

## Quotient rule

Quotient rule is applied on two functions to calculate the derivative which are in ratio.

{eq}u\left( x \right) {/eq} is the function in numerator and {eq}v\left( x \right) {/eq} in the denominator then the formula of quotient rule is

{eq}d\left( {\dfrac{{u\left( x \right)}}{{v\left( x \right)}}} \right) = \dfrac{{v\left( x \right)u'\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}} {/eq}

where {eq}u'\left( x \right) {/eq} means the derivative of function u.

## Answer and Explanation:

Rate of change of total receipts will be given by the derivative of a function {eq}T\left( x \right) {/eq}

{eq}\begin{align*} T'\left( x \right) &= \dfrac{d}{{dx}}\left( {\dfrac{{120}}{{{x^2} + 4}}} \right)\\ &= \dfrac{{\left( {{x^2} + 4} \right) \times \dfrac{d}{{dx}}\left( {120{x^2}} \right) - 120{x^2} \times \dfrac{d}{{dx}}\left( {{x^2} + 4} \right)}}{{{{\left( {{x^2} + 4} \right)}^2}}}\\ &= \dfrac{{960x}}{{{{\left( {{x^2} + 4} \right)}^2}}} \end{align*} {/eq}

1)

{eq}\begin{align*} T'\left( 1 \right) &= \dfrac{{960}}{{{{\left( {1 + 4} \right)}^2}}}\\ &= 38.40\;\rm{million} \end{align*} {/eq}

2)

{eq}\begin{align*} T'\left( 5 \right) &= \dfrac{{960 \times 5}}{{{{29}^2}}}\\ &= 5.707\;\rm{million} \end{align*} {/eq}

3)

{eq}\begin{align*} T'\left( 8 \right) &= \dfrac{{960 \times 8}}{{{{\left( {64 + 4} \right)}^2}}}\\ &= 1.66\;\rm{million} \end{align*} {/eq}

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