
What is the derivative graph of a parabola?
Answer
513.9k+ 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 number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

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

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

