Question

How do you find the derivative of $y={{(3x+1)}^{2}}$?

Answer 1

Hint: We will differentiate the given expression using chain rule. Firstly, we will consider \[3x+1\] as one function and differentiate the power to which the function \[3x+1\] is raised. Next, we will differentiate the function \[3x+1\], where \[3x\], which has \[x\] component, and 1, which is a constant are differentiated separately and we get derivatives of the above function.

Complete step by step solution:
According to the given question, we have to differentiate the function \[y\] with respect to \[x\], that is, we have to find the value of \[\dfrac{dy}{dx}\].
Chain rule states that if a function is written as \[f(g(x))\], then it derivative with respect to \[x\] can be given as,
\[\dfrac{d}{dx}(f(g(x)))=f'(g(x)).g'(x)\]
Applying the chain rule in the expression we get,
\[y={{(3x+1)}^{2}}\]
\[\Rightarrow \dfrac{d}{dx}(y)=\dfrac{d}{dx}({{(3x+1)}^{2}})\]
RHS part will get differentiated similar to the differentiation of \[{{x}^{2}}\], (that is, \[\dfrac{d}{dx}({{x}^{2}})=2x\])
So, we get,
\[\Rightarrow \dfrac{dy}{dx}=2(3x+1).\dfrac{d}{dx}(3x+1)\]
Now, we will differentiate the base that is, \[3x+1\], \[3x\] and 1 will be differentiated separately and so we get,
 \[\Rightarrow \dfrac{dy}{dx}=2(3x+1).(\dfrac{d}{dx}(3x)+\dfrac{d}{dx}(1))\]
We know that, derivative of x with respect to x Is 1 whereas derivative of a constant (such as 1) will be zero, we get,
\[\Rightarrow \dfrac{dy}{dx}=2(3x+1).(3\dfrac{d}{dx}(x)+0)\]
\[\Rightarrow \dfrac{dy}{dx}=2(3x+1).(3)\]
Multiplying the terms, we get,
\[\Rightarrow \dfrac{dy}{dx}=6(3x+1)=18x+6\]
Therefore, \[\dfrac{dy}{dx}=18x+6\].

Note: The chain rule proceeds progressively from the outermost function to the fundamental variable. The derivative of the above function can also be done by opening the parenthesis using the formula, \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and then differentiating the each term with respect to x, so we have,
\[y={{(3x+1)}^{2}}\]
Applying the formula, we get
\[y={{(3x)}^{2}}+2(3x)(1)+{{1}^{2}}\]
We get our expression as,
\[\Rightarrow y=9{{x}^{2}}+6x+1\]
Now, we will differentiate this expression with respect to x, then we get,
\[\dfrac{d}{dx}(y)=\dfrac{d}{dx}(9{{x}^{2}}+6x+1)\]
\[\Rightarrow \dfrac{d}{dx}(y)=\dfrac{d}{dx}(9{{x}^{2}})+\dfrac{d}{dx}(6x)+\dfrac{d}{dx}(1)\]
As derivative of a constant is zero, we get,
\[\Rightarrow \dfrac{dy}{dx}=9\dfrac{d}{dx}({{x}^{2}})+6\dfrac{d}{dx}(x)+0\]
\[\Rightarrow \dfrac{dy}{dx}=9(2x)+6\]
Rearranging we get the final expression as,
\[\Rightarrow \dfrac{dy}{dx}=18x+6\]
Therefore, \[\dfrac{dy}{dx}=18x+6\]