
How do you use the definition of a derivative to find the derivative of $f(x) = 2{x^2} + 5x?$
Answer
521.1k+ 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 solution:
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.
Take the given function –
$f(x) = 2{x^2} + 5x$
Therefore, $f'(x) = 2\dfrac{d}{{dx}}({x^2}) + 5\dfrac{d}{{dx}}(x)$
Using identity - $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$
$f'(x) = 2.2(x) + 5(1)$
Simplify the above expression –
$f'(x) = 4x + 5$
This is the required solution.
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.
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.
Complete step by step solution:
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.
Take the given function –
$f(x) = 2{x^2} + 5x$
Therefore, $f'(x) = 2\dfrac{d}{{dx}}({x^2}) + 5\dfrac{d}{{dx}}(x)$
Using identity - $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$
$f'(x) = 2.2(x) + 5(1)$
Simplify the above expression –
$f'(x) = 4x + 5$
This is the required solution.
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.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

