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Question: What is the derivative of \(f\left( x \right)={{x}^{3}}-3x\) ?...

What is the derivative of f(x)=x33xf\left( x \right)={{x}^{3}}-3x ?

Explanation

Solution

To find the derivative of the given function f(x)=x33xf\left( x \right)={{x}^{3}}-3x, we are going to use the following derivative form: dxndx=nxn1\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}. And this derivative form we are going to apply first on x3{{x}^{3}} and then on xx. Also, we are going to use the derivative form which says: d(kg(x))dx=kdg(x)dx\dfrac{d\left( kg\left( x \right) \right)}{dx}=k\dfrac{dg\left( x \right)}{dx}.

Complete step-by-step solution:
The function given in the above problem which we are going to take the derivative is as follows:
f(x)=x33xf\left( x \right)={{x}^{3}}-3x
Now, taking derivative with respect to x on both the sides of the above equation we get,
df(x)dx=d(x33x)dx\dfrac{df\left( x \right)}{dx}=\dfrac{d\left( {{x}^{3}}-3x \right)}{dx}
Distributing derivative amongst the two functions x3&3x{{x}^{3}}\And 3x we get,
df(x)dx=dx3dxd(3x)dx\dfrac{df\left( x \right)}{dx}=\dfrac{d{{x}^{3}}}{dx}-\dfrac{d\left( 3x \right)}{dx} ………………. (1)
Now, to differentiate the above function, we are going to use the following derivative property which states that:
dxndx=nxn1\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}
Applying the above property on dx3dx\dfrac{d{{x}^{3}}}{dx} by substituting the value of n as 3 in the above equation and we get,
dx3dx=3x31 dx3dx=3x2 \begin{aligned} & \dfrac{d{{x}^{3}}}{dx}=3{{x}^{3-1}} \\\ & \Rightarrow \dfrac{d{{x}^{3}}}{dx}=3{{x}^{2}} \\\ \end{aligned}
From the above, we have found the derivative of one of the part of the two derivatives so substituting the above value in eq. (1) we get,
df(x)dx=3x2d(3x)dx\dfrac{df\left( x \right)}{dx}=3{{x}^{2}}-\dfrac{d\left( 3x \right)}{dx} ………… (2)
Now, we are going to apply the following derivative form:
d(kg(x))dx=kdg(x)dx\dfrac{d\left( kg\left( x \right) \right)}{dx}=k\dfrac{dg\left( x \right)}{dx}
Substituting the value of k as 3 and g(x)=xg\left( x \right)=x in the above equation we get,
d(3x)dx=3dxdx\dfrac{d\left( 3x \right)}{dx}=3\dfrac{dx}{dx}
In the R.H.S of the above equation, dxdx will get cancelled out from the numerator and the denominator and we get 1 in place of dxdx\dfrac{dx}{dx} and we get,
d(3x)dx=3(1) d(3x)dx=3 \begin{aligned} & \dfrac{d\left( 3x \right)}{dx}=3\left( 1 \right) \\\ & \Rightarrow \dfrac{d\left( 3x \right)}{dx}=3 \\\ \end{aligned}
The above derivative is the second part of eq. (1) so using the above relation in eq. (2) we get,
df(x)dx=3x23\dfrac{df\left( x \right)}{dx}=3{{x}^{2}}-3
Now, taking 3 as common in the R.H.S of the above equation we get,
df(x)dx=3(x21)\dfrac{df\left( x \right)}{dx}=3\left( {{x}^{2}}-1 \right)
Hence, we have found the derivative of the given function as: 3(x21)3\left( {{x}^{2}}-1 \right).

Note: To solve the above problem you should know the following derivative forms otherwise you could not solve the problem further:
dxndx=nxn1; d(kg(x))dx=kdg(x)dx \begin{aligned} & \dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}; \\\ & \dfrac{d\left( kg\left( x \right) \right)}{dx}=k\dfrac{dg\left( x \right)}{dx} \\\ \end{aligned}
So, make sure you have properly understood the above derivatives.