What is the derivative graph of a parabola?
Answer
533.4k+ views
Hint: In this problem we need to find the derivative graph of a parabola. For this we will first assume the standard equation of the parabola which is given by $y=a{{x}^{2}}+bx+c$ where $a$, $b$, $c$ are the constants. Now we will differentiate the above equation with respect to the variable $x$. By using the differentiation formula, we will simplify the obtained equation to get the required result.
Complete step-by-step answer:
Let the equation of the parabola will be $y=a{{x}^{2}}+bx+c$.
Differentiating the above equation with respect to $x$, then we will get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}}+bx+c \right)$
Applying the differentiation for each term individually, then we will have
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}} \right)+\dfrac{d}{dx}\left( bx \right)+\dfrac{d}{dx}\left( c \right)$
Taking out the constants from differentiation which are in multiplication with the variables in the above equation, then we will get
$\dfrac{dy}{dx}=a\dfrac{d}{dx}\left( {{x}^{2}} \right)+b\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( c \right)$
From the differentiation formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ we can write the value of $\dfrac{d}{dx}\left( {{x}^{2}} \right)$ as $2x$. Substituting this value in the above equation, then we will have
$\dfrac{dy}{dx}=a\left( 2x \right)+b\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( c \right)$
We have the value $\dfrac{d}{dx}\left( x \right)=1$. Substituting this value in the above equation, then we will get
$\dfrac{dy}{dx}=a\left( 2x \right)+b\left( 1 \right)+\dfrac{d}{dx}\left( c \right)$
The value $c$ which is in the above equation is a constant. We know that differentiation value of a constant is equals to zero. By using this value, we can write the above equation as
$\dfrac{dy}{dx}=a\left( 2x \right)+b\left( 1 \right)+0$
Simplifying the above equation by using the basic mathematical operations, then we will get
$\dfrac{dy}{dx}=2ax+b$
The equation $2ax+b$ represents the equation of the line. We can observe this in the below graph also
Hence the derivative graph of the parabola is Straight Line.
Note: In this problem we have assumed the equation of the parabola as $y=a{{x}^{2}}+bx+c$ instead of $x=a{{y}^{2}}+by+c$ even though the both the equations represent the parabola. Because the equation $y=a{{x}^{2}}+bx+c$ has an extra advantage to calculate the differentiation value easily over the equation $x=a{{y}^{2}}+by+c$.
Complete step-by-step answer:
Let the equation of the parabola will be $y=a{{x}^{2}}+bx+c$.
Differentiating the above equation with respect to $x$, then we will get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}}+bx+c \right)$
Applying the differentiation for each term individually, then we will have
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}} \right)+\dfrac{d}{dx}\left( bx \right)+\dfrac{d}{dx}\left( c \right)$
Taking out the constants from differentiation which are in multiplication with the variables in the above equation, then we will get
$\dfrac{dy}{dx}=a\dfrac{d}{dx}\left( {{x}^{2}} \right)+b\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( c \right)$
From the differentiation formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ we can write the value of $\dfrac{d}{dx}\left( {{x}^{2}} \right)$ as $2x$. Substituting this value in the above equation, then we will have
$\dfrac{dy}{dx}=a\left( 2x \right)+b\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( c \right)$
We have the value $\dfrac{d}{dx}\left( x \right)=1$. Substituting this value in the above equation, then we will get
$\dfrac{dy}{dx}=a\left( 2x \right)+b\left( 1 \right)+\dfrac{d}{dx}\left( c \right)$
The value $c$ which is in the above equation is a constant. We know that differentiation value of a constant is equals to zero. By using this value, we can write the above equation as
$\dfrac{dy}{dx}=a\left( 2x \right)+b\left( 1 \right)+0$
Simplifying the above equation by using the basic mathematical operations, then we will get
$\dfrac{dy}{dx}=2ax+b$
The equation $2ax+b$ represents the equation of the line. We can observe this in the below graph also
Hence the derivative graph of the parabola is Straight Line.
Note: In this problem we have assumed the equation of the parabola as $y=a{{x}^{2}}+bx+c$ instead of $x=a{{y}^{2}}+by+c$ even though the both the equations represent the parabola. Because the equation $y=a{{x}^{2}}+bx+c$ has an extra advantage to calculate the differentiation value easily over the equation $x=a{{y}^{2}}+by+c$.
Recently Updated Pages
The magnetic field in a plane electromagnetic wave class 11 physics CBSE

In a plane electromagnetic wave the electric field class 12 physics CBSE

A plane electromagnetic wave travels in vacuum along class 12 physics CBSE

Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

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

What are the examples of C3 and C4 plants class 11 biology CBSE

State and prove Bernoullis theorem class 11 physics CBSE

10 examples of friction in our daily life

A body is said to be in dynamic equilibrium if A When class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

