Answer
Verified
485.7k+ views
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.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which places in India experience sunrise first and class 9 social science CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE