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  1. Asked: October 28, 2023In:

    In mathematics, a __________ is used to display the overlap and distinctions among sets.

    acedstud
    acedstud

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elemenRead more

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elements unique to each set. The circles are enclosed inside a rectangle which represents the universal set.

    For example, consider the universal set containing positive integers less than 10, and consider two sets: Set A, which contains even numbers, and Set B, which contains multiples of 3.

    A Venn diagram for these sets would have one circle representing even numbers and another circle representing multiples of 3. The overlapping section would represent numbers that are both even and multiples of 3, which in this case is 6. The other numbers that are neither even numbers nor multiples of 3 are placed within the rectangle but outside the circles.

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  2. Asked: January 10, 2021In:

    Differentiate y=x^{2}(x^{3}-1)^{3} wrt x

    acedstud
    acedstud

    To differentiate the function $y = x^2(x^3 - 1)^3$ with respect to $x$, we can use the product rule. The product rule states that if you have a function of the form $u(x)v(x)$, the derivative is given by: $$\dfrac{d}{dx}[u(x)v(x)] = u(x)v'(x) + u'(x)v(x)$$ In this case, let $u(x) = x^2$ and $v(x) =Read more

    To differentiate the function y = x^2(x^3 - 1)^3 with respect to x, we can use the product rule. The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

        \[\dfrac{d}{dx}[u(x)v(x)] = u(x)v'(x) + u'(x)v(x)\]

    In this case, let u(x) = x^2 and v(x) = (x^3 - 1)^3. We need to find their derivatives:

    1. u'(x) = \frac{d}{dx}(x^2) = 2x (using the power rule).

    2. We can find v'(x) using the chain rule. The chain rule states that if you have a composite function, such as v\left(w(x)\right), the derivative of v\left(w(x)\right) with respect to x is given by:

        \[\dfrac{d}{dx}\left[v\left(w(x)\right)\right] = v'\left(w(x)\right) * w'(x)\]

    To find v'(x), let w(x) = x^3 - 1, so v(x) = w(x)^3:

        \[v'(x) &= \frac{d}{dx}[w(x)^3] \\ &= 3w(x)^2 \cdot \frac{d}{dx}[w(x)]\]

    Now, we need to find \frac{d}{dx}[w(x)]:

        \[\frac{d}{dx}[w(x)] &= \frac{d}{dx}(x^3 - 1) \\ &= 3x^2\]

    Now, we can compute v'(x):

        \[v'(x) &= 3w(x)^2 \cdot 3x^2 \\ &= 9x^2(x^3 - 1)^2\]

    Now, apply the product rule to find \frac{d}{dx}[x^2(x^3 - 1)^3]:

        \[\frac{d}{dx}[x^2(x^3 - 1)^3] &= u(x)v'(x) + u'(x)v(x) \\ &= x^2 \cdot 9x^2(x^3 - 1)^2 + 2x \cdot (x^3 - 1)^3\]

    Now, simplify this expression:

        \[\frac{d}{dx}[x^2(x^3 - 1)^3] &= 9x^4(x^3 - 1)^2 + 2x(x^3 - 1)^3\]

    So, the derivative of y = x^2(x^3 - 1)^3 with respect to x is:

        \[\frac{d}{dx}[x^2(x^3 - 1)^3] &= 9x^4(x^3 - 1)^2 + 2x(x^3 - 1)^3\]

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  3. Asked: January 10, 2021In:

    Differentiate y=x(x^{2}-1)^{3} wrt x

    acedstud
    acedstud
    This answer was edited.

    To differentiate the function $y = x(x^2 - 1)^3$ with respect to $x$, you can apply both the product rule and the chain rule. Using the product rule, which states that $\dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x)$, we have: Let $u(x) = x$ and $v(x) = (x^2 - 1)^3$. We know that $u'(x)Read more

    To differentiate the function y = x(x^2 - 1)^3 with respect to x, you can apply both the product rule and the chain rule.

    Using the product rule, which states that \dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x), we have:

    Let u(x) = x and v(x) = (x^2 - 1)^3.

    We know that u'(x) = 1 (the derivative of x is 1).

    Now, we need to find v'(x), which involves applying the chain rule because v(x) is a composite function. The chain rule states that if you have a composite function, such as v\left(w(x)\right), the derivative of v\left(w(x)\right) with respect to x is given by:

    \dfrac{d}{dx}\left[v\left(w(x)\right)\right] = v'\left(w(x)\right) * w'(x).

    Let w(x) = x^2 - 1, so v(x) = w(x)^3.

    Now, we can find v'(x):

        \[v'(x) &= \dfrac{d}{dx}[w(x)^3] \\ &= 3w(x)^2 \cdot \dfrac{d}{dx}[w(x)]\]

    Now, we find \dfrac{d}{dx}[w(x)]:

        \[\dfrac{d}{dx}[w(x)] &= \dfrac{d}{dx}[x^2 - 1] \\ &= 2x\]

    Now, we can compute v'(x):

        \[v'(x) &= 3w(x)^2 \cdot 2x \\ &= 6x(x^2 - 1)^2\]

    Now, apply the product rule:

        \[\dfrac{d}{dx}[x(x^2 - 1)^3] &= u(x)v'(x) + u'(x)v(x) \\ &= x \cdot 6x(x^2 - 1)^2 + 1 \cdot (x^2 - 1)^3\]

    Now, simplify this expression:

        \[\dfrac{d}{dx}[x(x^2 - 1)^3] &= 6x^2(x^2 - 1)^2 + (x^2 - 1)^3\]

    So, the correct derivative of y = x(x^2 - 1)^3 with respect to x is:

        \[\dfrac{d}{dx}[x(x^2 - 1)^3] &= 6x^2(x^2 - 1)^2 + (x^2 - 1)^3\]

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  4. Asked: January 10, 2021In:

    Differentiate y=(2x-1)(x+4)^{2} wrt x

    acedstud
    acedstud

    To differentiate the function $y = (2x - 1)(x + 4)^2$ with respect to $x$, you can use the product rule. The product rule states that if you have a function of the form $u(x)v(x)$, the derivative is given by: $\dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x)$. In this case, let $u(x) = 2xRead more

    To differentiate the function y = (2x - 1)(x + 4)^2 with respect to x, you can use the product rule. The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

    \dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x).

    In this case, let u(x) = 2x - 1 and v(x) = (x + 4)^2.

    Now, we’ll find their derivatives:

    u'(x) = \dfrac{d}{dx} (2x - 1) = 2 (the derivative of 2x is 2, and the derivative of a constant, -1, is 0).

    To find v'(x), you can use the chain rule. The chain rule states that if you have a composite function, such as v\left(w(x)\right), the derivative of v\left(w(x)\right) with respect to x is given by:

    \dfrac{d}{dx}\left[v\left(w(x)\right)\right] = v'\left(w(x)\right) * w'(x).

    Let w(x) = x + 4, so v(x) = w(x)^2. Now, we can find v'(x):

    v'(x) = \dfrac{d}{dx} \left[w(x)^2\right] = 2w(x)w'(x).

    w'(x) = \dfrac{d}{dx} (x + 4) = 1 (the derivative of x is 1, and the derivative of a constant, 4, is 0).

    So, v'(x) = 2(x + 4) * 1 = 2(x + 4).

    Now, apply the product rule:

    \dfrac{d}{dx}\left[(2x - 1)(x + 4)^2\right] = u(x)v'(x) + u'(x)v(x) = (2x - 1) * 2(x + 4) + 2 * (x + 4)^2.

    Now, simplify this expression:

    (4x - 2)(x + 4) + 2(x + 4)^2.

    You can further simplify this expression if needed, but this is the derivative of y = (2x - 1)(x + 4)^2 with respect to x:

    y'(x) = (4x - 2)(x + 4) + 2(x + 4)^2.

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  5. Asked: January 10, 2021In:

    Differentiate y=3x^{3}(x^{2}+4)^{2} wrt x

    acedstud
    acedstud

    To differentiate the function $y = 3x^3(x^2 + 4)^2$ with respect to $x$, you can use the product rule. The product rule states that if you have a function of the form $u(x)v(x)$, the derivative is given by: $\dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x)$. In this case, let $u(x) = 3x^3$Read more

    To differentiate the function y = 3x^3(x^2 + 4)^2 with respect to x, you can use the product rule. The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

    \dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x).

    In this case, let u(x) = 3x^3 and v(x) = (x^2 + 4)^2.

    Now, we’ll find their derivatives:

    u'(x) = \dfrac{d}{dx} (3x^3) = 9x^2 (using the power rule).

    To find v'(x), you can use the chain rule. The chain rule states that if you have a composite function, such as v\left(w(x)\right), the derivative of v\left(w(x)\right) with respect to x is given by:

    \dfrac{d}{dx}\left[v\left(w(x)\right)\right] = v'\left(w(x)\right) * w'(x).

    Let w(x) = x^2 + 4, so v(x) = w(x)^2.

    Now, we can find v'(x):

    v'(x) = \dfrac{d}{dx} \left[w(x)^2\right] = 2w(x)w'(x).

    w'(x) = \dfrac{d}{dx} (x^2 + 4) = 2x (the derivative of x^2 is 2x, and the derivative of a constant, 4, is 0).

    So, v'(x) = 2(x^2 + 4) * 2x = 4x(x^2 + 4).

    Now, apply the product rule:

    \dfrac{d}{dx}\left[3x^3(x^2 + 4)^2\right] = u(x)v'(x) + u'(x)v(x) = 3x^3 * 4x(x^2 + 4) + 9x^2(x^2 + 4)^2.

    Now, simplify this expression:

    12x^4(x^2 + 4) + 9x^2(x^2 + 4)^2.

    So, the derivative of y = 3x^3(x^2 + 4)^2 with respect to x is:

    y'(x) = 12x^4(x^2 + 4) + 9x^2(x^2 + 4)^2.

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  6. Asked: January 10, 2021In:

    Differentiate y=(x^{2}+5)(3x-1)^{3} wrt x

    acedstud
    acedstud
    This answer was edited.

    To differentiate the function $y = (x^2 + 5)(3x - 1)^3$ with respect to $x$, you can use the product rule. The product rule states that if you have a function of the form $u(x)v(x)$, the derivative is given by: $\dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x)$. In this case, let $u(x) = xRead more

    To differentiate the function y = (x^2 + 5)(3x - 1)^3 with respect to x, you can use the product rule. The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

    \dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x).

    In this case, let u(x) = x^2 + 5 and v(x) = (3x - 1)^3.

    Now, we’ll find their derivatives:

    u'(x) = \dfrac{d}{dx} (x^2 + 5) = 2x (using the power rule).

    To find v'(x), we’ll need to use the chain rule. The chain rule states that if you have a composite function, such as v\left(w(x)\right), the derivative of v\left(w(x)\right) with respect to x is given by:

    \dfrac{d}{dx}\left[v\left(w(x)\right)\right] = v'\left(w(x)\right) * w'(x).

    Let w(x) = 3x - 1, so v(x) = w(x)^3.

    Now, we can find v'(x):

    v'(x) = \dfrac{d}{dx} \left[w(x)^3\right] = 3w(x)^2 * w'(x).

    w'(x) = \dfrac{d}{dx} (3x - 1) = 3 (the derivative of 3x is 3, and the derivative of a constant, -1, is 0).

    So, v'(x) = 3(3x - 1)^2 * 3 = 9(3x - 1)^2.

    Now, apply the product rule:

    \dfrac{d}{dx}\left[(x^2 + 5)(3x - 1)^3\right] = u(x)v'(x) + u'(x)v(x) = (x^2 + 5) * 9(3x - 1)^2 + 2x * (3x - 1)^3.

    Now, simplify this expression:

    9(x^2 + 5)(3x - 1)^2 + 2x(3x - 1)^3.

    So, the derivative of y = (x^2 + 5)(3x - 1)^3 with respect to x is:

    y'(x) = 9(x^2 + 5)(3x - 1)^2 + 2x(3x - 1)^3.

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  7. Asked: January 10, 2021In:

    Differentiate y=x^{2}(2x-5)^{4} wrt x

    acedstud
    acedstud
    This answer was edited.

    To differentiate the function $y = x^2(2x - 5)^4$ with respect to $x$, you use the product rule. The product rule states that if you have a function of the form $u(x)v(x)$, the derivative is given by: $\dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x)$. In this case, let $u(x) = x^2$ and $vRead more

    To differentiate the function y = x^2(2x - 5)^4 with respect to x, you use the product rule.

    The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

    \dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x).

    In this case, let u(x) = x^2 and v(x) = (2x - 5)^4.

    Now, we’ll find their derivatives:

    u'(x) = \dfrac{d}{dx} (x^2) = 2x (using the power rule).

    To find v'(x), we’ll need to use the chain rule. The chain rule states that if you have a composite function, such as v\left(w(x)\right), the derivative of v\left(w(x)\right) with respect to x is given by:

    \dfrac{d}{dx}\left[v\left(w(x)\right)\right] = v'\left(w(x)\right) * w'(x).

    Let w(x) = 2x - 5, so v(x) = w(x)^4. Now, we can find v'(x):

    v'(x) = \dfrac{d}{dx} \left[w(x)^4\right] = 4w(x)^3 * w'(x).

    w'(x) = \dfrac{d}{dx} (2x - 5) = 2 (the derivative of 2x is 2, and the derivative of a constant, -5, is 0).

    So,

    v'(x) = 4(2x - 5)^3 * 2 = 8(2x - 5)^3.

    Now, we can apply the product rule:

    \dfrac{d}{dx}\left[x^2(2x - 5)^4\right] = u(x)v'(x) + u'(x)v(x) = x^2 * 8(2x - 5)^3 + 2x * (2x - 5)^4.

    Now, simplify this expression:

    8x^2(2x - 5)^3 + 2x(2x - 5)^4.

    So, the corrected derivative of y = x^2(2x - 5)^4 with respect to x is:

    y'(x) = 8x^2(2x - 5)^3 + 2x(2x - 5)^4.

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  8. Asked: January 10, 2021In:

    Differentiate y=(3x-2)(x^{2}+3) wrt x

    acedstud
    acedstud

    To differentiate the function $y = (3x - 2)(x^2 + 3)$ with respect to $x$, you can use the product rule, which states that the derivative of a product of two functions is given by: $\frac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x)$, where $u(x)$ and $v(x)$ are functions of $x$, and $u'(x)$Read more

    To differentiate the function y = (3x - 2)(x^2 + 3) with respect to x, you can use the product rule, which states that the derivative of a product of two functions is given by:

    \frac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x),

    where u(x) and v(x) are functions of x, and u'(x) and v'(x) are their respective derivatives with respect to x.

    In this case, let u(x) = 3x - 2 and v(x) = x^2 + 3.

    Now, let’s find their derivatives:

    u'(x) = \dfrac{d}{dx} (3x - 2) = 3, (the derivative of 3x is 3, and the derivative of a constant, -2, is 0).

    v'(x) = \dfrac{d}{dx} (x^2 + 3) = 2x, (use the power rule to find the derivative of x^2, which is 2x, and the derivative of a constant, 3, is 0).

    Now, apply the product rule:

    \dfrac{d}{dx}\left[(3x - 2)(x^2 + 3)\right] = (3x - 2)(2x) + (3)(x^2 + 3).

    Now, simplify this expression:

    (3x - 2)(2x) + 3(x^2 + 3) = 6x^2 - 4x + 3x^2 + 9.

    Combine like terms:

    6x^2 + 3x^2 - 4x + 9 = 9x^2 - 4x + 9.

    So, the derivative of y = (3x - 2)(x^2 + 3) with respect to x is:

    y'(x) = 9x^2 - 4x + 9.

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  9. Asked: October 25, 2023In:

    A __________ illustrates how sets intersect and differ from one another.

    acedstud
    acedstud

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elemenRead more

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elements unique to each set. The circles are enclosed inside a rectangle which represents the universal set.

    For example, consider the universal set containing positive integers less than 10, and consider two sets: Set A, which contains even numbers, and Set B, which contains multiples of 3.

    A Venn diagram for these sets would have one circle representing even numbers and another circle representing multiples of 3. The overlapping section would represent numbers that are both even and multiples of 3, which in this case is 6. The other numbers that are neither even numbers nor multiples of 3 are placed within the rectangle but outside the circles.

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  10. Asked: October 25, 2023In:

    A __________(Venn diagram, line plot) shows the relationship between sets.

    acedstud
    acedstud
    This answer was edited.

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elemenRead more

    A Venn diagram is a visual tool used to represent the relationship between sets. It consists of one or more circles that overlap or intersect, with each circle representing a specific set. The overlapping regions show where sets have elements in common, while the non-overlapping parts display elements unique to each set. The circles are enclosed inside a rectangle which represents the universal set.

    For example, consider the universal set containing positive integers less than 10, and consider two sets: Set A, which contains even numbers, and Set B, which contains multiples of 3.

    A Venn diagram for these sets would have one circle representing even numbers and another circle representing multiples of 3. The overlapping section would represent numbers that are both even and multiples of 3, which in this case is 6. The other numbers that are neither even numbers nor multiples of 3 are placed within the rectangle but outside the circles.

    See less