
How do you use the definition of a derivative to find the derivative of $f(x) = 2{x^2} + 5x?$
Answer
473.1k+ 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.
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