# If f(t) = (t^2 + 7 t + 3) (4 t^2 + 3), find f '(t).

## Question:

If {eq}\displaystyle f(t) = (t^2 + 7 t + 3) (4 t^2 + 3) {/eq}, find {eq}\displaystyle f '(t) {/eq}.

## Product Rule of Differentiation:

If we have two functions {eq}a(x) {/eq} and {eq}b(x) {/eq}, then:

{eq}\frac{d}{{dx}}\left( {\left( {a(x)} \right)\left( {b(x)} \right)} \right) = \left( {b\left( x \right)\frac{d}{{dx}}\left( {a\left( x \right)} \right) + a\left( x \right)\frac{d}{{dx}}\left( {b\left( x \right)} \right)} \right) {/eq}

Using this formula, we can solve many derivative up to simplification.

## Answer and Explanation:

Here, in this case, the given derivative can be solved as:

{eq}\eqalign{ f'(t)& = \frac{d}{{dt}}\left[ {\left( {{t^2} + 7t + 3} \right)\left( {4{t^2} + 3} \right)} \right] \cr & = \left( {{t^2} + 7t + 3} \right)\frac{d}{{dt}}\left( {4{t^2} + 3} \right) + \left( {4{t^2} + 3} \right)\frac{d}{{dt}}\left( {{t^2} + 7t + 3} \right)\,\,\,\left[ {{\text{Using product rule}}} \right] \cr & = \left( {{t^2} + 7t + 3} \right)\left( {16t} \right) + \left( {4{t^2} + 3} \right)\left( {2{t^2} + 7} \right) \cr & = 8{t^4} + 16{t^3} + 146{t^2} + 48t + 21 \cr} {/eq}

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from Saxon Calculus Homeschool: Online Textbook Help

Chapter 20 / Lesson 1