Question

Differentiate.

\(F\left( x \right) = \frac{1}{{2{x^3} - 6{x^2} + 5}}\)

\(\)

Short answer

Derivative of the function is \( - \frac{{6{x^2} - 12x}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\).

Step-by-step solution

Find the derivative of f using the Quotient rule

The equation for the Quotient rule is \({\left( {\frac{f}{g}} \right)^\prime } = \frac{{gf' - fg'}}{{{{\left( g \right)}^2}}}\).

Apply Quotient rule for the function.

\(\begin{aligned}{\frac{d}{{dx}}\left( {F\left( x \right)} \right)} &= \frac{d}{{dx}}\left( {\frac{1}{{2{x^3} - 6{x^2} + 5}}} \right)\\ &= \frac{{\left( {2{x^3} - 6{x^2} + 5} \right)\frac{d}{{dx}}\left( 1 \right) - \left( 1 \right)\frac{d}{{dx}}\left( {2{x^3} - 6{x^2} + 5} \right)}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\end{aligned}\)

Simplify the obtained condition

Above derivative rule and find the answer.

\(\begin{aligned}F'\left( x \right) &= \frac{{\left( {2{x^3} - 6{x^2} + 5} \right)\frac{d}{{dx}}\left( 1 \right) - \left( 1 \right)\frac{d}{{dx}}\left( {2{x^3} - 6{x^2} + 5} \right)}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\\ &= \frac{{\left( {2{x^3} - 6{x^2} + 5} \right)\left( 0 \right) - \left( 1 \right)\left( {6{x^2} - 12x} \right)}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\\ &= - \frac{{6{x^2} - 12x}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\end{aligned}\)

So, \(F'\left( x \right) = - \frac{{6{x^2} - 12x}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\)is the final answer of this derivative function.