Answer
Verified
399.6k+ 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
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
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
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Who was the Governor general of India at the time of class 11 social science CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
Name five important trees found in the tropical evergreen class 10 social studies CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE