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Question: How do you find the derivative of \(y=\dfrac{5}{{{(x-3)}^{2}}}\) ?...

How do you find the derivative of y=5(x3)2y=\dfrac{5}{{{(x-3)}^{2}}} ?

Explanation

Solution

We start solving the given question y=5(x3)2y=\dfrac{5}{{{(x-3)}^{2}}} by using the chain rule dydx=(dydz)(dzdx) \dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right) in the given equation. Here, we can take z as x - 3. Then we can solve the derivatives separately. And finally multiply both the derivatives to get the final answer.

Complete step by step solution:
According to the problem, we are asked to find the derivative of the equation y=5(x3)2y=\dfrac{5}{{{(x-3)}^{2}}}--- ( 1 )
Therefore, by using the chain rule dydx=(dydz)(dzdx)\dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right), in equation 1, let us do the following:
Let us take z = x - 3
First, we calculate for dydz\dfrac{dy}{dz}. Now, let us put z = x – 3 in dydz\dfrac{dy}{dz}.
Therefore, dydz=d(5z2)dz\dfrac{dy}{dz}=\dfrac{d(\dfrac{5}{{{z}^{2}}})}{dz} [Since z = x - 3]
dydz=d(5.z2)dz\Rightarrow \dfrac{dy}{dz}=\dfrac{d(5.z^{-2})}{dz}
Let us use the formula - dxndx=nxn1 \dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}. So, by doing this, we get the following.
dydz=(5)(2)z3\Rightarrow \dfrac{dy}{dz}=\left(5\right)\left(-2\right){{z}^{-3}}
dydz=10z3\Rightarrow \dfrac{dy}{dz}=-10{{z}^{-3}} ----(2)
Let us consider the above equation as 2.
Now, let us calculate for dzdx \dfrac{dz}{dx}.
Let us again put z = x - 3, in the above equation, that is dzdx \dfrac{dz}{dx}. Therefore, we get
dzdx=d(x3)dx\Rightarrow \dfrac{dz}{dx}=\dfrac{d\left(x-3\right)}{dx}
dzdx=d(x)dxd(3)dx\Rightarrow \dfrac{dz}{dx}=\dfrac{d\left(x\right)}{dx}-\dfrac{d\left(3\right)}{dx}
dzdx=10\Rightarrow \dfrac{dz}{dx}=1-0
dzdx=1\Rightarrow \dfrac{dz}{dx}=1 ----(3)
Since, we calculated both dzdx \dfrac{dz}{dx} and dydz \dfrac{dy}{dz}, let us substitute both the equations (2) and (3) in equation (1), that is dydx=(dydz)(dzdx) \dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right).
dydx=(10z3)(1)\Rightarrow \dfrac{dy}{dx}=(-10{{z}^{-3}})(1) ----(4)
But now, we should substitute, z = x - 3. Therefore, let us substitute equation (4) with z = x - 3.
dydx=(10(x3)3)\Rightarrow \dfrac{dy}{dx}=(-10{(x-3)}^{-3})
dydx=10(x3)3\Rightarrow \dfrac{dy}{dx}=\dfrac{-10}{{{(x-3)}^{3}}}
So, we have found the derivative of the given equation y=5(x3)2y=\dfrac{5}{{{(x-3)}^{2}}} as dydx=10(x3)3 \dfrac{dy}{dx}=\dfrac{-10}{{{(x-3)}^{3}}}
Therefore, the solution of the given equation y=5(x3)2y=\dfrac{5}{{{(x-3)}^{2}}} is dydx=10(x3)3 \dfrac{dy}{dx}=\dfrac{-10}{{{(x-3)}^{3}}}

Note:
Whenever we get this type of problem, we should always be careful while substituting the equations.Also, we could also find the derivative of the equation using the quotient rule which is y=f(x)g(x)f(x)g(x)(g(x))2y'=\dfrac{f'(x)g(x)-f(x)g'(x)}{{{(g(x))}^{2}}} . You could also check the answer by integrating the answer to get back the question.