
Find the derivative of the following function:
${x^{ - 3}}\left( {5 + 3x} \right)$
Answer
514.5k+ views
Hint: In this question apply the product rule of differentiation which is given as $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ later on apply the formula of differentiation of $\dfrac{d}{{dx}}\left( {{x^{ - n}}} \right) = \left( { - n} \right){{\text{x}}^{ - n - 1}}{\text{ and }}\dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0{\text{ and }}\dfrac{d}{{dx}}\left( {nx} \right) = n$ so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Let
$y = {x^{ - 3}}\left( {5 + 3x} \right)$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{x^{ - 3}}\left( {5 + 3x} \right)} \right]$
Now here we use product rule of differentiate which is given as
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {x^{ - 3}}\dfrac{d}{{dx}}\left( {5 + 3x} \right) + \left( {5 + 3x} \right)\dfrac{d}{{dx}}\left( {{x^{ - 3}}} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {{x^{ - n}}} \right) = \left( { - n} \right){{\text{x}}^{ - n - 1}}{\text{ and }}\dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0{\text{ and }}\dfrac{d}{{dx}}\left( {nx} \right) = n$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {x^{ - 3}}\left( {0 + 3} \right) + \left( {5 + 3x} \right)\left( { - 3{x^{ - 3 - 1}}} \right)$
Now simplify this equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = 3{x^{ - 3}} - 15{x^{ - 4}} - 9{x^{ - 3}}$
$ \Rightarrow \dfrac{d}{{dx}}y = - 15{x^{ - 4}} - 6{x^{ - 3}} = - 3{x^{ - 3}}\left( {5{x^{ - 1}} + 2} \right)$
So this is the required differentiation.
Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of ${x^n}$ differentiation which is stated above then first apply the product rule as above then use the property of differentiation of ${x^n}$ as above and simplify we will get the required answer.
Complete step-by-step answer:
Let
$y = {x^{ - 3}}\left( {5 + 3x} \right)$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{x^{ - 3}}\left( {5 + 3x} \right)} \right]$
Now here we use product rule of differentiate which is given as
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {x^{ - 3}}\dfrac{d}{{dx}}\left( {5 + 3x} \right) + \left( {5 + 3x} \right)\dfrac{d}{{dx}}\left( {{x^{ - 3}}} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {{x^{ - n}}} \right) = \left( { - n} \right){{\text{x}}^{ - n - 1}}{\text{ and }}\dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0{\text{ and }}\dfrac{d}{{dx}}\left( {nx} \right) = n$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {x^{ - 3}}\left( {0 + 3} \right) + \left( {5 + 3x} \right)\left( { - 3{x^{ - 3 - 1}}} \right)$
Now simplify this equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = 3{x^{ - 3}} - 15{x^{ - 4}} - 9{x^{ - 3}}$
$ \Rightarrow \dfrac{d}{{dx}}y = - 15{x^{ - 4}} - 6{x^{ - 3}} = - 3{x^{ - 3}}\left( {5{x^{ - 1}} + 2} \right)$
So this is the required differentiation.
Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of ${x^n}$ differentiation which is stated above then first apply the product rule as above then use the property of differentiation of ${x^n}$ as above and simplify we will get the required answer.
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