
Find the derivative of the following function:
${x^4}\left( {5\sin x - 3\cos x} \right)$
Answer
612.3k+ 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 in the solution apply the differentiation property of sin x, cos x and ${x^n}$ which is given as $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ and }}\dfrac{d}{{dx}}\left( {\cos x} \right) = -\sin x{\text{ and }}\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Let
$y = {x^4}\left( {5\sin x - 3\cos x} \right)$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{x^4}\left( {5\sin x - 3\cos x} \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^4}\dfrac{d}{{dx}}\left( {5\sin x - 3\cos x} \right) + \left( {5\sin x - 3\cos x} \right)\dfrac{d}{{dx}}{x^4}$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ and }}\dfrac{d}{{dx}}\left( {\cos x} \right) = -\sin x{\text{ and }}\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {x^4}\left( {5\cos x + 3\sin x} \right) + \left( {5\sin x - 3\cos x} \right)4{x^3}$
Now simplify this equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = 5{x^4}\cos x + 3{x^4}\sin x + 20{x^3}\sin x - 12{x^3}\cos x$
$ \Rightarrow \dfrac{d}{{dx}}y = {x^3}\cos x\left( {5x - 12} \right) + {x^3}\sin x\left( {3x + 20} \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 sin x, cos x and ${x^n}$ differentiation which is stated above then first apply the product rule as above then use the property of differentiation of sin x, cos x and ${x^n}$ as above and simplify we will get the required answer.
Complete step-by-step answer:
Let
$y = {x^4}\left( {5\sin x - 3\cos x} \right)$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{x^4}\left( {5\sin x - 3\cos x} \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^4}\dfrac{d}{{dx}}\left( {5\sin x - 3\cos x} \right) + \left( {5\sin x - 3\cos x} \right)\dfrac{d}{{dx}}{x^4}$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ and }}\dfrac{d}{{dx}}\left( {\cos x} \right) = -\sin x{\text{ and }}\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {x^4}\left( {5\cos x + 3\sin x} \right) + \left( {5\sin x - 3\cos x} \right)4{x^3}$
Now simplify this equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = 5{x^4}\cos x + 3{x^4}\sin x + 20{x^3}\sin x - 12{x^3}\cos x$
$ \Rightarrow \dfrac{d}{{dx}}y = {x^3}\cos x\left( {5x - 12} \right) + {x^3}\sin x\left( {3x + 20} \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 sin x, cos x and ${x^n}$ differentiation which is stated above then first apply the product rule as above then use the property of differentiation of sin x, cos x and ${x^n}$ as above and simplify we will get the required answer.
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