Question

# Find the derivative of the following:$-\dfrac{2651}{504\sqrt[315]{{{x}^{2966}}}}$

Hint: Donâ€™t get confused with the large numbers. Use the formula $\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{\dfrac{n}{m}}}} \right)=\dfrac{d}{dx}\left( {{x}^{-\dfrac{n}{m}}} \right)=\dfrac{-n}{m}{{x}^{\dfrac{-n}{m}-1}}$to find the derivative.

The given function is $-\dfrac{2651}{504\sqrt[315]{{{x}^{2966}}}}$.
Taking out the constant term, we get
$\dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{-2651}{504}\times \dfrac{d}{dx}\left\{ \dfrac{1}{\sqrt[315]{{{x}^{2966}}}} \right\}$
The can be written as,
$\dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{-2651}{504}\times \dfrac{d}{dx}\left\{ \dfrac{1}{{{x}^{\dfrac{2966}{315}}}} \right\}$
So the above equation can be re-written using the formula $\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{\dfrac{n}{m}}}} \right)=\dfrac{d}{dx}\left( {{x}^{-\dfrac{n}{m}}} \right)=\dfrac{-n}{m}{{x}^{\dfrac{-n}{m}-1}},$
\begin{align} & \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{-2651}{504}\times \dfrac{-2966}{315}\times {{x}^{\dfrac{-2966}{315}-1}} \\ & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{-2651}{504}\times \dfrac{-2966}{315}\times {{x}^{\dfrac{-2966-315}{315}}} \\ & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{-2651}{504}\times \dfrac{-2966}{315}\times {{x}^{\dfrac{-3281}{315}}} \\ & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 2966}{504\times 315}\times \dfrac{1}{\sqrt[315]{{{x}^{3281}}}} \\ \end{align}
Dividing throughout by â€˜2â€™, we get
$\Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{\sqrt[315]{{{x}^{3150+131}}}}$
Now applying the formula ${{x}^{m+n}}={{x}^{m}}.{{x}^{n}}$ under the root, we have
\begin{align} & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{\sqrt[315]{{{x}^{3150}}.{{x}^{131}}}} \\ & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{\sqrt[315]{{{x}^{3150}}}\times \sqrt[315]{{{x}^{131}}}} \\ \end{align}
Now, by applying the formula ${{x}^{mn}}={{({{x}^{m}})}^{n}}$ under the root, we get
\begin{align} & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{\sqrt[315]{{{({{x}^{10}})}^{315}}}\times \sqrt[315]{{{x}^{131}}}} \\ & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{{{x}^{10}}\times \sqrt[315]{{{x}^{131}}}} \\ \end{align}
Here we can observe that $(315-131=184)$, so we will rationalise by $\sqrt[315]{{{x}^{184}}}$, we get
\begin{align} & \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\dfrac{1}{{{x}^{10}}}\times \dfrac{1}{\sqrt[315]{{{x}^{131}}}}\times \dfrac{\sqrt[315]{{{x}^{184}}}}{\sqrt[315]{{{x}^{184}}}} \\ & \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{{{x}^{10}}}\times \dfrac{\sqrt[315]{{{x}^{184}}}}{\sqrt[315]{{{x}^{131}}\times {{x}^{184}}}} \\ & \Rightarrow \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{{{x}^{10}}}\dfrac{\sqrt[315]{{{x}^{184}}}}{\sqrt[315]{{{x}^{315}}}} \\ & \dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\times \dfrac{1}{{{x}^{10}}}\dfrac{\sqrt[315]{{{x}^{184}}}}{x} \\ \end{align}
$\dfrac{d}{dx}\left\{ \dfrac{-2651}{504\sqrt[315]{{{x}^{2966}}}} \right\}=\dfrac{2651\times 1483}{252\times 315}\dfrac{\sqrt[315]{{{x}^{184}}}}{{{x}^{11}}}$

Note: In this problem looking at the bigger values the student may get confused. We should always try to solve the problem using a simple basic formula. The student sometimes gets confused to remove the constant term while deriving and will make mistakes.