Question

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

Answer 1

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, $\dfrac{dy}{dx}=18x+6$