Question

derivative of f(x)=(2x + 1)^3

Answer 1

Use the chain rule:

\[f'(x) = 3(2x+1)^2*2\]

So , take derivative considering the thing inside the parenteses as a n "x" and then multiply by the derivative of the inside thing, in this case 2

Answer 2

First of all, I have : power = 3 and x = 2x +1 so :
\(3(2x+1)^2\)

Now, differentiate (2x+1 ) = 2 + 0 = 2 so I have :

\(3(2)(2x+1)^2\) = \(6(2x+1)^2

Answer 3

Let it be : \(f(x) = X^3\) where \( X = 2x + 1\)

So : differentiation of \(X^3 = 3X^2\)

Now differentiate X : \(X = 2x + 1\) So , derivative of 2x + 1 (which is equal to X) ,

\(2 + 0\) = \(2\)

So I will write this as : \(3X^2 * 2 = 6X^2 = 6(2x+1)^2\)

Answer 4

@Christos , got it?

Answer 5

Formula :

Let : \(y = ( f(x) )^n \)

\(\boxed{\large{\cfrac{dy}{dx} = n (f(x))^{n-1} * \cfrac{d(f(x))}{dx} }}\)

Rule used : Chain rule.

Answer 6

@Christos , you there?

Answer 7

If you're unable to get this method :

I am giving you another question just like yours one.


\(\LARGE{\textbf{Example : }}\) \(\large{\textbf{Differentiate : } \quad (3x+1)^2 }\)

\(\LARGE{\textbf{Solution : }} \) Now, let , \(\large{ y = (3x+1)^2 }\)

As per the formula I have :

\(\boxed{\large{\cfrac{dy}{dx} = n (f(x))^{n-1} * \cfrac{d(f(x))}{dx} }}\)

So :


\(\large{\cfrac{dy}{dx} = 2 ( 3x+1 ) ^{2 - 1} * \cfrac{d(3x+1)}{dx}} \\
\implies \cfrac{dy}{dx } = 2(3x+1) * (3x^{1-1} + 0 ) \\
\implies \cfrac{dy}{dx} = 2(3x+1) * 3 \\
\\
\\
\boxed{\cfrac{dy}{dx} = 6(3x+1) } ---- [\textbf{Answer} ] \)

Answer 8

I got it thanks!

Answer 9

Great to know, @Christos . Best of luck.