Find The Derivative Of The Given Function.$\[ F(x) = 4x^6 - 2x^4 + X^3 - 4x + 5 \\]$\[ F'(x) = \\] $\[ \square \\]

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Introduction

In calculus, the derivative of a function represents the rate of change of the function with respect to one of its variables. It is a fundamental concept in mathematics and has numerous applications in various fields, including physics, engineering, and economics. In this article, we will focus on finding the derivative of a given polynomial function.

What is a Polynomial Function?

A polynomial function is a function that can be written in the form:

f(x)=anxn+an−1xn−1+…+a1x+a0{ f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 }

where an,an−1,…,a1,a0a_n, a_{n-1}, \ldots, a_1, a_0 are constants, and nn is a non-negative integer. The degree of the polynomial is the highest power of xx in the function.

The Power Rule

One of the most important rules in finding the derivative of a polynomial function is the power rule. The power rule states that if f(x)=xnf(x) = x^n, then f′(x)=nxn−1f'(x) = nx^{n-1}.

Finding the Derivative of the Given Function

Now, let's apply the power rule to find the derivative of the given function:

f(x)=4x6−2x4+x3−4x+5{ f(x) = 4x^6 - 2x^4 + x^3 - 4x + 5 }

To find the derivative, we will differentiate each term separately.

Differentiating the First Term

The first term is 4x64x^6. Using the power rule, we have:

ddx(4x6)=4⋅6x6−1=24x5{ \frac{d}{dx}(4x^6) = 4 \cdot 6x^{6-1} = 24x^5 }

Differentiating the Second Term

The second term is −2x4-2x^4. Using the power rule, we have:

ddx(−2x4)=−2⋅4x4−1=−8x3{ \frac{d}{dx}(-2x^4) = -2 \cdot 4x^{4-1} = -8x^3 }

Differentiating the Third Term

The third term is x3x^3. Using the power rule, we have:

ddx(x3)=3x3−1=3x2{ \frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2 }

Differentiating the Fourth Term

The fourth term is −4x-4x. Using the power rule, we have:

ddx(−4x)=−4⋅1x1−1=−4{ \frac{d}{dx}(-4x) = -4 \cdot 1x^{1-1} = -4 }

Differentiating the Fifth Term

The fifth term is 55. Since the derivative of a constant is zero, we have:

ddx(5)=0{ \frac{d}{dx}(5) = 0 }

Combining the Derivatives

Now, let's combine the derivatives of each term to find the derivative of the given function:

f′(x)=24x5−8x3+3x2−4{ f'(x) = 24x^5 - 8x^3 + 3x^2 - 4 }

Conclusion

In this article, we have found the derivative of a given polynomial function using the power rule. We have differentiated each term separately and combined the derivatives to find the final derivative. The derivative of the given function is:

f′(x)=24x5−8x3+3x2−4{ f'(x) = 24x^5 - 8x^3 + 3x^2 - 4 }

This derivative can be used to find the rate of change of the function with respect to xx.

Example Problems

Problem 1

Find the derivative of the function f(x)=3x4−2x2+x−1f(x) = 3x^4 - 2x^2 + x - 1.

Solution

Using the power rule, we have:

ddx(3x4)=3⋅4x4−1=12x3{ \frac{d}{dx}(3x^4) = 3 \cdot 4x^{4-1} = 12x^3 }

ddx(−2x2)=−2⋅2x2−1=−4x{ \frac{d}{dx}(-2x^2) = -2 \cdot 2x^{2-1} = -4x }

ddx(x)=1x1−1=1{ \frac{d}{dx}(x) = 1x^{1-1} = 1 }

ddx(−1)=0{ \frac{d}{dx}(-1) = 0 }

Combining the derivatives, we have:

f′(x)=12x3−4x+1{ f'(x) = 12x^3 - 4x + 1 }

Problem 2

Find the derivative of the function f(x)=2x3+3x2−4x+1f(x) = 2x^3 + 3x^2 - 4x + 1.

Solution

Using the power rule, we have:

ddx(2x3)=2⋅3x3−1=6x2{ \frac{d}{dx}(2x^3) = 2 \cdot 3x^{3-1} = 6x^2 }

ddx(3x2)=3⋅2x2−1=6x{ \frac{d}{dx}(3x^2) = 3 \cdot 2x^{2-1} = 6x }

ddx(−4x)=−4⋅1x1−1=−4{ \frac{d}{dx}(-4x) = -4 \cdot 1x^{1-1} = -4 }

ddx(1)=0{ \frac{d}{dx}(1) = 0 }

Combining the derivatives, we have:

f′(x)=6x2+6x−4{ f'(x) = 6x^2 + 6x - 4 }

Practice Problems

Problem 1

Find the derivative of the function f(x)=x4−2x2+3x−1f(x) = x^4 - 2x^2 + 3x - 1.

Problem 2

Find the derivative of the function f(x)=2x3+x2−3x+1f(x) = 2x^3 + x^2 - 3x + 1.

Problem 3

Find the derivative of the function f(x)=x5−4x3+2x2−3x+1f(x) = x^5 - 4x^3 + 2x^2 - 3x + 1.

Problem 4

Find the derivative of the function f(x)=3x4−2x3+x2−4x+1f(x) = 3x^4 - 2x^3 + x^2 - 4x + 1.

Problem 5

Find the derivative of the function f(x)=2x3+3x2−4x+1f(x) = 2x^3 + 3x^2 - 4x + 1.

Solutions

Problem 1

Using the power rule, we have:

ddx(x4)=4x4−1=4x3{ \frac{d}{dx}(x^4) = 4x^{4-1} = 4x^3 }

ddx(−2x2)=−2⋅2x2−1=−4x{ \frac{d}{dx}(-2x^2) = -2 \cdot 2x^{2-1} = -4x }

ddx(3x)=3⋅1x1−1=3{ \frac{d}{dx}(3x) = 3 \cdot 1x^{1-1} = 3 }

ddx(−1)=0{ \frac{d}{dx}(-1) = 0 }

Combining the derivatives, we have:

f′(x)=4x3−4x+3{ f'(x) = 4x^3 - 4x + 3 }

Problem 2

Using the power rule, we have:

ddx(2x3)=2⋅3x3−1=6x2{ \frac{d}{dx}(2x^3) = 2 \cdot 3x^{3-1} = 6x^2 }

ddx(3x2)=3⋅2x2−1=6x{ \frac{d}{dx}(3x^2) = 3 \cdot 2x^{2-1} = 6x }

ddx(−3x)=−3⋅1x1−1=−3{ \frac{d}{dx}(-3x) = -3 \cdot 1x^{1-1} = -3 }

ddx(1)=0{ \frac{d}{dx}(1) = 0 }

Combining the derivatives, we have:

f′(x)=6x2+6x−3{ f'(x) = 6x^2 + 6x - 3 }

Problem 3

Using the power rule, we have:

ddx(x5)=5x5−1=5x4{ \frac{d}{dx}(x^5) = 5x^{5-1} = 5x^4 }

ddx(−4x3)=−4⋅3x3−1=−12x2{ \frac{d}{dx}(-4x^3) = -4 \cdot 3x^{3-1} = -12x^2 }

ddx(2x2)=2⋅2x2−1=4x{ \frac{d}{dx}(2x^2) = 2 \cdot 2x^{2-1} = 4x }

ddx(−3x)=−3⋅1x1−1=−3{ \frac{d}{dx}(-3x) = -3 \cdot 1x^{1-1} = -3 }

ddx(1)=0{ \frac{d}{dx}(1) = 0 }

Combining the derivatives, we have:

f′(x)=5x4−12x2+4x−3{ f'(x) = 5x^4 - 12x^2 + 4x - 3 }

Problem 4

Using the power rule, we have:

ddx(3x4)=3⋅4x4−1=12x3{ \frac{d}{dx}(3x^4) = 3 \cdot 4x^{4-1} = 12x^3 }

ddx(−2x3)=−2⋅3x3−1=−6x2{ \frac{d}{dx}(-2x^3) = -2 \cdot 3x^{3-1} = -6x^2 }

ddx(x2)=2x2−1=2x{ \frac{d}{dx}(x^2) = 2x^{2-1} = 2x }

ddx(−4x)=−4⋅1x1−1=−4{ \frac{d}{dx}(-4x) = -4 \cdot 1x^{1-1} = -4 }

ddx(1)=0{ \frac{d}{dx}(1) = 0 }

Combining the derivatives, we have:

f′(x)=12x3−6x2+2x−4{ f'(x) = 12x^3 - 6x^2 + 2x - 4 }

Problem 5

Using the power rule, we have:

${ \frac{d}{dx}(2

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Introduction

In our previous article, we discussed how to find the derivative of a polynomial function using the power rule. In this article, we will answer some frequently asked questions about the derivative of a polynomial function.

Q: What is the derivative of a constant function?

A: The derivative of a constant function is zero. This is because the derivative represents the rate of change of the function with respect to one of its variables, and a constant function does not change with respect to its variable.

Q: How do I find the derivative of a polynomial function with a negative exponent?

A: To find the derivative of a polynomial function with a negative exponent, you can use the power rule. For example, if you have a function of the form f(x)=ax−nf(x) = ax^{-n}, then the derivative is f′(x)=−nax−n−1f'(x) = -nax^{-n-1}.

Q: Can I use the power rule to find the derivative of a polynomial function with a fractional exponent?

A: Yes, you can use the power rule to find the derivative of a polynomial function with a fractional exponent. For example, if you have a function of the form f(x)=ax1nf(x) = ax^{\frac{1}{n}}, then the derivative is f′(x)=1nax1n−1f'(x) = \frac{1}{n}ax^{\frac{1}{n}-1}.

Q: How do I find the derivative of a polynomial function with multiple terms?

A: To find the derivative of a polynomial function with multiple terms, you can use the power rule for each term separately. For example, if you have a function of the form f(x)=axm+bxn+cxpf(x) = ax^m + bx^n + cx^p, then the derivative is f′(x)=maxm−1+nbxn−1+pcxp−1f'(x) = max^{m-1} + nbx^{n-1} + pcx^{p-1}.

Q: Can I use the power rule to find the derivative of a polynomial function with a variable in the exponent?

A: Yes, you can use the power rule to find the derivative of a polynomial function with a variable in the exponent. For example, if you have a function of the form f(x)=axbxf(x) = ax^{bx}, then the derivative is f′(x)=abxbx−1xbxf'(x) = abx^{bx-1}x^{bx}.

Q: How do I find the derivative of a polynomial function with a trigonometric function in the exponent?

A: To find the derivative of a polynomial function with a trigonometric function in the exponent, you can use the chain rule and the power rule. For example, if you have a function of the form f(x)=axsin⁡xf(x) = ax^{\sin x}, then the derivative is f′(x)=acos⁡xxsin⁡x−1f'(x) = a\cos x x^{\sin x - 1}.

Q: Can I use the power rule to find the derivative of a polynomial function with a logarithmic function in the exponent?

A: Yes, you can use the power rule to find the derivative of a polynomial function with a logarithmic function in the exponent. For example, if you have a function of the form f(x)=axlog⁡xf(x) = ax^{\log x}, then the derivative is f′(x)=a1xxlog⁡x−1f'(x) = a\frac{1}{x}x^{\log x - 1}.

Q: How do I find the derivative of a polynomial function with a product of two functions?

A: To find the derivative of a polynomial function with a product of two functions, you can use the product rule. For example, if you have a function of the form f(x)=axm⋅bxnf(x) = ax^m \cdot bx^n, then the derivative is f′(x)=maxm−1bxn+abxm⋅nbxn−1f'(x) = max^{m-1}bx^n + abx^m \cdot nbx^{n-1}.

Q: Can I use the power rule to find the derivative of a polynomial function with a quotient of two functions?

A: Yes, you can use the quotient rule to find the derivative of a polynomial function with a quotient of two functions. For example, if you have a function of the form f(x)=axmbxnf(x) = \frac{ax^m}{bx^n}, then the derivative is f′(x)=maxm−1bxn−abxm⋅nbxn−1(bxn)2f'(x) = \frac{max^{m-1}bx^n - abx^m \cdot nbx^{n-1}}{(bx^n)^2}.

Conclusion

In this article, we have answered some frequently asked questions about the derivative of a polynomial function. We have discussed how to find the derivative of a polynomial function with a negative exponent, a fractional exponent, multiple terms, a variable in the exponent, a trigonometric function in the exponent, a logarithmic function in the exponent, a product of two functions, and a quotient of two functions. We hope that this article has been helpful in understanding the derivative of a polynomial function.

Practice Problems

Problem 1

Find the derivative of the function f(x)=3x4−2x2+x−1f(x) = 3x^4 - 2x^2 + x - 1.

Problem 2

Find the derivative of the function f(x)=2x3+3x2−4x+1f(x) = 2x^3 + 3x^2 - 4x + 1.

Problem 3

Find the derivative of the function f(x)=x5−4x3+2x2−3x+1f(x) = x^5 - 4x^3 + 2x^2 - 3x + 1.

Problem 4

Find the derivative of the function f(x)=3x4−2x3+x2−4x+1f(x) = 3x^4 - 2x^3 + x^2 - 4x + 1.

Problem 5

Find the derivative of the function f(x)=2x3+3x2−4x+1f(x) = 2x^3 + 3x^2 - 4x + 1.

Solutions

Problem 1

Using the power rule, we have:

[ \frac{d}{dx}(3x^4) = 3 \cdot 4x^{4-1} = 12x^3 }$

ddx(−2x2)=−2⋅2x2−1=−4x{ \frac{d}{dx}(-2x^2) = -2 \cdot 2x^{2-1} = -4x }

ddx(x)=1x1−1=1{ \frac{d}{dx}(x) = 1x^{1-1} = 1 }

ddx(−1)=0{ \frac{d}{dx}(-1) = 0 }

Combining the derivatives, we have:

f′(x)=12x3−4x+1{ f'(x) = 12x^3 - 4x + 1 }

Problem 2

Using the power rule, we have:

ddx(2x3)=2⋅3x3−1=6x2{ \frac{d}{dx}(2x^3) = 2 \cdot 3x^{3-1} = 6x^2 }

ddx(3x2)=3⋅2x2−1=6x{ \frac{d}{dx}(3x^2) = 3 \cdot 2x^{2-1} = 6x }

ddx(−4x)=−4⋅1x1−1=−4{ \frac{d}{dx}(-4x) = -4 \cdot 1x^{1-1} = -4 }

ddx(1)=0{ \frac{d}{dx}(1) = 0 }

Combining the derivatives, we have:

f′(x)=6x2+6x−4{ f'(x) = 6x^2 + 6x - 4 }

Problem 3

Using the power rule, we have:

ddx(x5)=5x5−1=5x4{ \frac{d}{dx}(x^5) = 5x^{5-1} = 5x^4 }

ddx(−4x3)=−4⋅3x3−1=−12x2{ \frac{d}{dx}(-4x^3) = -4 \cdot 3x^{3-1} = -12x^2 }

ddx(2x2)=2⋅2x2−1=4x{ \frac{d}{dx}(2x^2) = 2 \cdot 2x^{2-1} = 4x }

ddx(−3x)=−3⋅1x1−1=−3{ \frac{d}{dx}(-3x) = -3 \cdot 1x^{1-1} = -3 }

ddx(1)=0{ \frac{d}{dx}(1) = 0 }

Combining the derivatives, we have:

f′(x)=5x4−12x2+4x−3{ f'(x) = 5x^4 - 12x^2 + 4x - 3 }

Problem 4

Using the power rule, we have:

ddx(3x4)=3⋅4x4−1=12x3{ \frac{d}{dx}(3x^4) = 3 \cdot 4x^{4-1} = 12x^3 }

ddx(−2x3)=−2⋅3x3−1=−6x2{ \frac{d}{dx}(-2x^3) = -2 \cdot 3x^{3-1} = -6x^2 }

ddx(x2)=2x2−1=2x{ \frac{d}{dx}(x^2) = 2x^{2-1} = 2x }

ddx(−4x)=−4⋅1x1−1=−4{ \frac{d}{dx}(-4x) = -4 \cdot 1x^{1-1} = -4 }

ddx(1)=0{ \frac{d}{dx}(1) = 0 }

Combining the derivatives, we have:

f′(x)=12x3−6x2+2x−4{ f'(x) = 12x^3 - 6x^2 + 2x - 4 }

Problem 5

Using the power rule, we have:

ddx(2x3)=2⋅3x3−1=6x2{ \frac{d}{dx}(2x^3) = 2 \cdot 3x^{3-1} = 6x^2 }

ddx(3x2)=3⋅2x2−1=6x{ \frac{d}{dx}(3x^2) = 3 \cdot 2x^{2-1} = 6x }

[ \frac{d}{dx}(-4x) = -4 \cdot 1x^{1-1} = -4 \