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Question: For some constants \( a \) and \( b \) , find the derivative of \( {{\left( ax+b \right)}^{2}} \) ....

For some constants aa and bb , find the derivative of (ax+b)2{{\left( ax+b \right)}^{2}} .

Explanation

Solution

Hint : We are going to find the derivative (ax+b)2{{\left( ax+b \right)}^{2}} by chain rule. The expression (ax+b)2{{\left( ax+b \right)}^{2}} is in the form of xn{{x}^{n}} where n=2n=2 so first of all we are differentiating (ax+b)2{{\left( ax+b \right)}^{2}} with respect to x in the same way as we differentiate x2{{x}^{2}} . We know the differentiation of x2{{x}^{2}} as 2x. Now, substitute ax+bax+b in place of x so we get the differentiation as 2(ax+b)2\left( ax+b \right) . Now, multiply the derivative of ax+bax+b with 2(ax+b)2\left( ax+b \right) and hence, you will get the derivative.

Complete step-by-step answer :
We have to find the derivative of (ax+b)2{{\left( ax+b \right)}^{2}} where a and b are constants.
We are going to find the derivative of (ax+b)2{{\left( ax+b \right)}^{2}} by chain rule.
The formula for chain rule is given by:
dydx=d(f(g(x)))dx=f′(g(x))×g′(x)\dfrac{dy}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right) ……… Eq. (1)
Comparing the f(g(x))&g(x)f\left( g\left( x \right) \right)\And g\left( x \right) with the given expression (ax+b)2{{\left( ax+b \right)}^{2}} we get,
f(g(x))=(ax+b)2 g(x)=ax+b \begin{aligned} & f\left( g\left( x \right) \right)={{\left( ax+b \right)}^{2}} \\\ & g\left( x \right)=ax+b \\\ \end{aligned}
To differentiate (ax+b)2{{\left( ax+b \right)}^{2}} by chain rule we have to find the above derivatives and then multiply them.
We are going to find the derivative of f(g(x))=(ax+b)2f\left( g\left( x \right) \right)={{\left( ax+b \right)}^{2}} in which as you can see that (ax+b)2{{\left( ax+b \right)}^{2}} is in the form of xn{{x}^{n}} so first of all we are assuming that ax+bax+b as x and differentiating the given expression.
We know that derivative of xn{{x}^{n}} with respect to x is:
dxndx=nxn−1\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}
The value of n=2n=2 so differentiating (ax+b)2{{\left( ax+b \right)}^{2}} with respect to x by keeping ax+bax+b as it is we get,
f′(g(x))=d(ax+b)2dx=2(ax+b){f}'\left( g\left( x \right) \right)=\dfrac{d{{\left( ax+b \right)}^{2}}}{dx}=2\left( ax+b \right) ………. Eq. (2)
Differentiating g(x)=ax+bg\left( x \right)=ax+b with respect to x we get,
g′(x)=d(ax+b)dx=a{g}'\left( x \right)=\dfrac{d\left( ax+b \right)}{dx}=a …….. Eq. (3)
Substituting the values from eq. (2) and eq. (3) in eq. (1) we get,
d(ax+b)2dx=2(ax+b)a\dfrac{d{{\left( ax+b \right)}^{2}}}{dx}=2\left( ax+b \right)a
Rearranging the above equation we get,
d(ax+b)2dx=2a(ax+b)\dfrac{d{{\left( ax+b \right)}^{2}}}{dx}=2a\left( ax+b \right)

Note : Here, the student should apply every formula correctly to get the correct derivative of the given function quickly. The place of blunder here is that you might wrongly do the derivative of the expression written in the bracket i.e. ax+bax+b so be careful while differentiating it.