Question

Find the derivatives of 1(x^3+sin)^2

Answer 1

Answer:

2(x³ + sin(x))(3x² + cos(x))

Step-by-step explanation:

As per the information provided in the question, We have :

• y = 1(x^3+sin)^2

we are asked to find d/dx with respect to y.

In order to find the derivative, We will use chain rule.

It is given by –

$\begin{gathered}\longmapsto \rm \dfrac{d}{dx}(y) = f'(g(x)) \times \frac{d}{dx} g(x) \end{gathered}$

By applying the rule,

$\begin{gathered}\longmapsto \rm \dfrac{d}{dx}(y) = {x}^{3} + \sin(x) \frac{d}{dx} ( {x}^{3} + \sin(x) ) \times 2 \end{gathered}$

Derivating the secondary function,

$\begin{gathered}\longmapsto \rm \dfrac{d}{dx}(y) =2 ( {x}^{3} + \sin(x)) ( 3 \times {x}^{3 - 1} + \cos(x) ) \end{gathered}$

$\begin{gathered}\longmapsto \rm \dfrac{d}{dx}(y) =2 ( {x}^{3} + \sin(x)) ( 3 {x}^{2} + \cos(x) ) \end{gathered}$

∴ The derivative of 1(x^3+sin)^2 is 2(x³ + sin(x))(3x² + cos(x)).

Learn more!

$\begin{gathered}\boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\text{Differentiation \:formulae:}}\\\\\bigstar\:\:\sf{ \dfrac{d}{dx} \times {{x}^{n}}=n\times{x}^{n-1}}\\\\\bigstar\:\:\sf{ \dfrac{d}{dx}({x})=1}\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( { Cons}) =0\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {log(x)}) =\dfrac{1}{x} \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sqrt{x}} )=\dfrac{1}{2 \sqrt{x}}\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\sin(x)})=\cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\cos(x)})=-\sin(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\sin(x)})=\cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\tan(x)})={\sec}^{2}(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\cot(x)})=-{ \cosec}^{2}(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\sec(x)})=\sec(x)\times \tan(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\cosec(x)})= - \cosec(x)\times\cot(x)\end{array}}\end{gathered}$