Answer
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Hint: Don’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.
Complete step by step solution:
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 ‘2’, 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.
Complete step by step solution:
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 ‘2’, 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.
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