Question

How do you find the derivative of $-2x{{\left( {{x}^{2}}+3 \right)}^{-2}}$?

Answer 1

Hint: We first define the multiplication rule and how the differentiation of function works. We take addition of these two different differentiated values. We take the $\dfrac{dy}{dx}$ altogether. We keep one function and differentiate the other one and then do the same thing with the other function. Then we take the addition to complete the formula.

Complete step by step solution:
We now discuss the multiplication process of two functions where \[f\left( x \right)=u\left( x \right)v\left( x \right)\]
Differentiating \[f\left( x \right)=uv\], we get \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\].
The above-mentioned rule is the multiplication rule. We apply that on $f\left( x \right)=-2x{{\left( {{x}^{2}}+3 \right)}^{-2}}$. We assume the functions where \[u\left( x \right)=-2x,v\left( x \right)={{\left( {{x}^{2}}+3 \right)}^{-2}}\]
We know that differentiation of \[u\left( x \right)=-2x\] is ${{u}^{'}}\left( x \right)=-2$.
Now we use chain rule for the differentiation of \[v\left( x \right)={{\left( {{x}^{2}}+3 \right)}^{-2}}\] .
We use the rule of \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\].
We get \[{{v}^{'}}\left( x \right)=\left[ \left( -2 \right){{\left( {{x}^{2}}+3 \right)}^{-3}} \right]\times \left( 2x \right)=-4x{{\left( {{x}^{2}}+3 \right)}^{-3}}\].
We now take differentiation on both parts of $f\left( x \right)=-2x{{\left( {{x}^{2}}+3 \right)}^{-2}}$ and get \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ -2x{{\left( {{x}^{2}}+3 \right)}^{-2}} \right]\]
We place the values of ${{u}^{'}}\left( x \right)=-2$ and \[{{v}^{'}}\left( x \right)=-4x{{\left( {{x}^{2}}+3 \right)}^{-3}}\] to get
\[\dfrac{d}{dx}\left[ -2x{{\left( {{x}^{2}}+3 \right)}^{-2}} \right]=\left( -2x \right)\dfrac{d}{dx}\left[ {{\left( {{x}^{2}}+3 \right)}^{-2}} \right]+\left[ {{\left( {{x}^{2}}+3 \right)}^{-2}} \right]\dfrac{d}{dx}\left( -2x \right)\].
We take all the $\dfrac{dy}{dx}$ forms altogether to get
\[\begin{align}
  & \dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ -2x{{\left( {{x}^{2}}+3 \right)}^{-2}} \right] \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\left( -2x \right)\left[ -4x{{\left( {{x}^{2}}+3 \right)}^{-3}} \right]+\left[ {{\left( {{x}^{2}}+3 \right)}^{-2}} \right]\left( -2 \right) \\
 & \Rightarrow {{f}^{'}}\left( x \right)=2{{\left( {{x}^{2}}+3 \right)}^{-2}}\left( \dfrac{4{{x}^{2}}}{{{x}^{2}}+3}-1 \right) \\
\end{align}\]
Therefore, differentiation of $-2x{{\left( {{x}^{2}}+3 \right)}^{-2}}$ is \[2{{\left( {{x}^{2}}+3 \right)}^{-2}}\left( \dfrac{4{{x}^{2}}}{{{x}^{2}}+3}-1 \right)\].

Note: We need remember that in the chain rule \[\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}\], we aren’t cancelling out the part \[d\left[ h\left( x \right) \right]\]. Cancelation of the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate. The rule may be extended or generalized to many other situations, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts.