
What is the derivative limit of $y=mx+b$?
Answer
529.8k+ views
Hint: To obtain the derivative limit of a given equation use limit definition for derivatives which involves the function. Firstly we will write the general formula to find the derivative limit of the function. Then we will substitute the required value in the formula by taking help of the function and finally we will simplify the obtained equation and get the desired answer.
Complete step-by-step answer:
The function given in the question is as below:
$y=mx+b$….$\left( 1 \right)$
The formula for derivative limit of a function is given as below:
$\dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{f\left( x+c \right)-f\left( x \right)}{c}$….$\left( 2 \right)$
On comparing equation (1) and equation (2) we get,
$y\left( x \right)=f\left( x \right)$
$\therefore f\left( x \right)=mx+b$….$\left( 3 \right)$
So value of other function is as below:
$\begin{align}
& f\left( x+c \right)=y\left( x+c \right) \\
& \Rightarrow f\left( x+c \right)=m\left( x+c \right)+b \\
\end{align}$
$\therefore f\left( x+c \right)=mx+mc+b$…..$\left( 4 \right)$
Substitute value from equation (3) and (4) in equation (2) and simplify to get,
$\begin{align}
& \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{\left( mx+mc+b \right)-\left( mx+b \right)}{c} \\
& \Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{mx+mc+b-mx-b}{c} \\
& \Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{mc}{c} \\
& \Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}m \\
& \therefore \dfrac{dy}{dx}=m \\
\end{align}$
Hence derivative limit of $y=mx+b$ is $m$
Note: Limit is a value that a function approaches when value is put in it. Limit is widely used in fields of mathematics such as in continuity, integral and also derivatives. We can understand the limit role in finding a derivative by the mean of the graph also. We can check whether our answer is correct or not by simply differentiating the function in a normal way which is a less complicated and easily solved method as it doesn't contain any concept of limit or continuity. But using the method of limit definition is a very fundamental way to find the solution of calculus problems. Derivatives are the rate at which the quantity changes instantaneously to another quantity.
Complete step-by-step answer:
The function given in the question is as below:
$y=mx+b$….$\left( 1 \right)$
The formula for derivative limit of a function is given as below:
$\dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{f\left( x+c \right)-f\left( x \right)}{c}$….$\left( 2 \right)$
On comparing equation (1) and equation (2) we get,
$y\left( x \right)=f\left( x \right)$
$\therefore f\left( x \right)=mx+b$….$\left( 3 \right)$
So value of other function is as below:
$\begin{align}
& f\left( x+c \right)=y\left( x+c \right) \\
& \Rightarrow f\left( x+c \right)=m\left( x+c \right)+b \\
\end{align}$
$\therefore f\left( x+c \right)=mx+mc+b$…..$\left( 4 \right)$
Substitute value from equation (3) and (4) in equation (2) and simplify to get,
$\begin{align}
& \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{\left( mx+mc+b \right)-\left( mx+b \right)}{c} \\
& \Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{mx+mc+b-mx-b}{c} \\
& \Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}\dfrac{mc}{c} \\
& \Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{c \to 0}m \\
& \therefore \dfrac{dy}{dx}=m \\
\end{align}$
Hence derivative limit of $y=mx+b$ is $m$
Note: Limit is a value that a function approaches when value is put in it. Limit is widely used in fields of mathematics such as in continuity, integral and also derivatives. We can understand the limit role in finding a derivative by the mean of the graph also. We can check whether our answer is correct or not by simply differentiating the function in a normal way which is a less complicated and easily solved method as it doesn't contain any concept of limit or continuity. But using the method of limit definition is a very fundamental way to find the solution of calculus problems. Derivatives are the rate at which the quantity changes instantaneously to another quantity.
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