How do you find the instantaneous rate of change of function at a point?
Answer
639.9k+ views
Hint: The instantaneous rate of change of function at a point can be defined as the derivative of the function and it can be expressed as $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
Complete step-by-step answer:
One can find the instantaneous rate of change of the function at a point by finding the derivative of that function and placing it in the x-value of the point.
Instantaneous rate of change of the function can be represented by the slope of the line, which says how much the function is increasing or decreasing as the x-values change. It can also be defined as $\dfrac{{dy}}{{dx}}$
The instantaneous rate of change can be expressed as the change in the rate at a particular instant and it is the same as the change in the derivative value at a specific point.
Let us compute the instantaneous rate of change of the function $f(x) = 3{x^2} + 12$ at the point $x = 4$
Take the given function –
$f(x) = 3{x^2} + 12$
Using identity - $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$
Therefore, $f'(x) = 6x$
Now, place the instantaneous rate of change at $x = 4$ in the above equation.
$f'(4) = 6(4)$
Simplify the above equation –
$f'(4) = 24$
This is the required solution.
Additional Information:
The difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverse of each other.
Note: Remember different identities to solve the given function and to find out the rate of change such as we did in the example with respect to “x”. Also, be good in basic mathematical expressions for the simplification.
Complete step-by-step answer:
One can find the instantaneous rate of change of the function at a point by finding the derivative of that function and placing it in the x-value of the point.
Instantaneous rate of change of the function can be represented by the slope of the line, which says how much the function is increasing or decreasing as the x-values change. It can also be defined as $\dfrac{{dy}}{{dx}}$
The instantaneous rate of change can be expressed as the change in the rate at a particular instant and it is the same as the change in the derivative value at a specific point.
Let us compute the instantaneous rate of change of the function $f(x) = 3{x^2} + 12$ at the point $x = 4$
Take the given function –
$f(x) = 3{x^2} + 12$
Using identity - $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$
Therefore, $f'(x) = 6x$
Now, place the instantaneous rate of change at $x = 4$ in the above equation.
$f'(4) = 6(4)$
Simplify the above equation –
$f'(4) = 24$
This is the required solution.
Additional Information:
The difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverse of each other.
Note: Remember different identities to solve the given function and to find out the rate of change such as we did in the example with respect to “x”. Also, be good in basic mathematical expressions for the simplification.
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