Find the derivative of the function f ( x ) = 4 x 2 − 9 x f(x)=4x^2-9x f ( x ) = 4 x 2 − 9 x .

f ′ ( x ) = 8 x − 9 f^{\prime}\left(x\right)=8x-9 f ′ ( x ) = 8 x − 9

f ′ ( x ) = 4 x − 18 f^{\prime}\left(x\right)=4x-18 f ′ ( x ) = 4 x − 18

f ′ ( x ) = 0 f^{\prime}\left(x\right)=0 f ′ ( x ) = 0

f ′ ( x ) = 4 x − 9 f^{\prime}\left(x\right)=4x-9 f ′ ( x ) = 4 x − 9

How do you find the derivative of the function f(x) = x 2 ?

To find the derivative of the function f ( x ) = x 2 , we use the limit definition of the derivative: f ′ ( x ) = lim h → 0 f ( x + h ) - f ( x ) h . For f ( x ) = x 2 , this becomes lim h → 0 ( x + h ) 2 - x 2 h . Simplifying, we get 2 x h + h 2 h , which further simplifies to 2 x + h . Taking the limit as h → 0 , we get 2 x . Thus, the derivative is f ′ ( x ) = 2 x .

What is the limit definition of a derivative?

The limit definition of a derivative is a fundamental concept in calculus that defines the derivative of a function at a point. It is given by the formula: f ′ ( x ) = lim h → 0 f ( x + h ) - f ( x ) h . This formula calculates the instantaneous rate of change of the function at any given point x by considering the limit as h approaches 0.

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Topic summary

Derivatives represent the slope of a tangent line at any point on a function. To find the derivative of a function, such as $f (x) = x^{2}$ , we use the limit definition: $f' (x) = lim_h \to 0 (f(x+h)-f(x))/h$ . For $x = 1$ , the slope is $2$ , and for $x = - 2$ , it is $- 4$ . This process allows for the calculation of slopes at any x-value efficiently.

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Derivatives

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Derivatives Video Summary

In calculus, the concept of a derivative is fundamental, representing the slope of a tangent line at a specific point on a function. To find the derivative of a function, we can use the limit definition of a derivative, which is expressed as:

\[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]

In this equation, $h$ represents a small change in $x$, and as $h$ approaches zero, we can determine the slope of the tangent line at any point $x$. For example, consider the function $f(x) = x^2$. To find its derivative, we substitute into the limit definition:

\[f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h}\]

Expanding $(x + h)^2$ gives us $x^2 + 2xh + h^2$. Substituting this back into the equation results in:

\[f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}\]

Factoring out $h$ from the numerator allows us to simplify:

\[f'(x) = \lim_{h \to 0} (2x + h)\]

As $h$ approaches zero, this simplifies to:

\[f'(x) = 2x\]

This result indicates that the derivative of $f(x) = x^2$ is $f'(x) = 2x$, which provides the slope of the tangent line at any point $x$. For specific values, such as $x = 1$ and $x = -2$, we can easily calculate:

For $x = 1$:

\[f'(1) = 2(1) = 2\]

For $x = -2$:

\[f'(-2) = 2(-2) = -4\]

Thus, the slope of the tangent line at $x = 1$ is 2, and at $x = -2$ it is -4. This method allows us to find the slope of the tangent line at any point on the curve defined by the function $f(x) = x^2$ simply by substituting the desired $x$ value into the derivative equation $f'(x) = 2x$. This approach can be applied to any differentiable function, making it a powerful tool in calculus.

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Problem

Find the derivative of the function $f(x)=4x^2-9x$ .

$f^{\prime}\left(x\right)=4x-18$

$f^{\prime}\left(x\right)=8x-9$

$f^{\prime}\left(x\right)=0$

$f^{\prime}\left(x\right)=4x-9$

Problem

Use the definition of a derivative, to find the derivative of the function $g(x)=x^3$ at $x=-1$ .

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The derivative of a function represents the slope of the tangent line at any point on the function. It is defined using the limit process as follows: $f_{'} (x) = lim_{h \to 0} \frac{f_{(x + h)} - f_{(x)}}{h}$ . This formula calculates the instantaneous rate of change of the function at any given point x.

To find the derivative of the function $f (x) = x^{2}$ , we use the limit definition of the derivative: $f_{'} (x) = lim_{h \to 0} \frac{f_{(x + h)} - f_{(x)}}{h}$ . For $f (x) = x^{2}$ , this becomes $lim_{h \to 0} \frac{(^{x + h) 2} - x^{2}}{h}$ . Simplifying, we get $\frac{2 x h + h^{2}}{h}$ , which further simplifies to $2 x + h$ . Taking the limit as $h \to 0$ , we get $2 x$ . Thus, the derivative is $f_{'} (x) = 2 x$ .

To find the slope of the tangent line at a specific point using the derivative, you first need the general derivative of the function. For example, if the function is $f (x) = x^{2}$ , the derivative is $f_{'} (x) = 2 x$ . To find the slope at a specific point, say $x = 1$ , you substitute $1$ into the derivative: $f_{'} (1) = 2 \times 1 = 2$ . Therefore, the slope of the tangent line at $x = 1$ is $2$ .

The limit definition of a derivative is a fundamental concept in calculus that defines the derivative of a function at a point. It is given by the formula: $f_{'} (x) = lim_{h \to 0} \frac{f_{(x + h)} - f_{(x)}}{h}$ . This formula calculates the instantaneous rate of change of the function at any given point x by considering the limit as h approaches 0.

To use the limit definition to find the derivative of a function, follow these steps: 1) Write the function in the form $f (x)$ . 2) Substitute $x + h$ into the function to get $f (x + h)$ . 3) Use the limit definition formula: $f_{'} (x) = lim_{h \to 0} \frac{f_{(x + h)} - f_{(x)}}{h}$ . 4) Simplify the expression inside the limit. 5) Take the limit as $h \to 0$ . This will give you the derivative of the function.