
What is the differentiation of \[uv\]?
Answer
498.3k+ views
Hint: Differentiation of \[uv\] can be obtained by the product rule. Product rule is the formula used to find the derivatives of a product of two or more functions. if \[u,v\] are two functions of \[x\], then the derivative of product of \[uv\] is given by \[\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)\].
Complete step-by-step solution:
From the question it is clear that we have to write the differentiation of \[uv\]. Which we can simply write from the product rule.
Let us assume \[u,v\] are two functions of \[x\].
Product rule is the formula used to find the derivatives of a product of two or more functions.
When the first function is multiplied by the derivative of the second function plus the second function multiplied by the derivative of the first function is known as product rule. This is an easy way of remembering and easy writing.
Let us assume \[uv\] as \[y\]
\[\Rightarrow y=uv\]……………(1)
From the question we were asked to find the derivative of \[uv\]. So, finding \[\dfrac{dy}{dx}\] is equal to finding \[\dfrac{d}{dx}\left( uv \right)\].
So,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( uv \right)\]……………….(2)
Using product rule, we can write the equation (2) as
\[\Rightarrow \dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)\]
So finally, we can conclude that derivative of \[uv\]is \[u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)\].
Note: Students should be careful while applying product rules. Small errors in the application of product rule may lead to doing this question wrong and may get the wrong answer. Many students may have misconception that \[\dfrac{d}{dx}\left( uv \right)=\dfrac{d}{dx}\left( u \right)\times \dfrac{d}{dx}\left( v \right)\], which is completely wrong formula. Using this wrong formula may lead to this question wrong.
Complete step-by-step solution:
From the question it is clear that we have to write the differentiation of \[uv\]. Which we can simply write from the product rule.
Let us assume \[u,v\] are two functions of \[x\].
Product rule is the formula used to find the derivatives of a product of two or more functions.
When the first function is multiplied by the derivative of the second function plus the second function multiplied by the derivative of the first function is known as product rule. This is an easy way of remembering and easy writing.
Let us assume \[uv\] as \[y\]
\[\Rightarrow y=uv\]……………(1)
From the question we were asked to find the derivative of \[uv\]. So, finding \[\dfrac{dy}{dx}\] is equal to finding \[\dfrac{d}{dx}\left( uv \right)\].
So,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( uv \right)\]……………….(2)
Using product rule, we can write the equation (2) as
\[\Rightarrow \dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)\]
So finally, we can conclude that derivative of \[uv\]is \[u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)\].
Note: Students should be careful while applying product rules. Small errors in the application of product rule may lead to doing this question wrong and may get the wrong answer. Many students may have misconception that \[\dfrac{d}{dx}\left( uv \right)=\dfrac{d}{dx}\left( u \right)\times \dfrac{d}{dx}\left( v \right)\], which is completely wrong formula. Using this wrong formula may lead to this question wrong.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

