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castortr0y [4]
3 years ago
9

How would you differentiate this problem using the quotient rule?

Mathematics
1 answer:
ioda3 years ago
6 0
Rewrite the square root of X as X^ 1/2. Then take the derivative. Since 1/3 is a constant with respect to X you can pull that out. So, you should have
1/3 d/dx [ (2x^2-x^5)/3x^1/2].

Then factor x^2 out of 2x^2-x^5.

Then by using the negative exponent rule you should have

1/3 d/dx [ x^2(2-x^3)x^-1/2 ]

Then multiply x^2 by x ^-1/2 by adding the exponents.

1/3 d/dx [ x^3/2 (2-x^3)]

Then differentiate.

You should get x^1/2 - (3x^7/2)/2.

I hope that helps.
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Find the area of a triangle whose vertices are located at (-2,5), (-4, -3), and (3,1). Evaluate the determinant using diagonals.
myrzilka [38]

Answer:

The area of the triangle is 48 unit²

Step-by-step explanation:

The given vertices of the triangle are;

(-2, 5), (-4, -3), and (3, 1)

The formula for finding the area of a triangle with given coordinates of the vertices is as follows;

\Delta  = \dfrac{1}{2}\times \begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix} = \dfrac{1}{2}\times \left | x_1\cdot y_2 - x_2\cdot y_1 + x_2\cdot y_3 - x_3\cdot y_2 + x_3\cdot y_1 - x_1\cdot y_3\right |

Substituting gives;

\Delta  = \dfrac{1}{2}\times \begin{vmatrix}-2& 5 & 1\\ -4 & -3 & 1\\ 3 & 1 & 1\end{vmatrix} \\\Delta  = \dfrac{1}{2}\times \left | (-2)\times (-3) - ((-4)\times 5) + (-4)\times 1 - 3\times (-3) + 3 \times 5 - (-2)\times 1\right | \\\Delta  = \dfrac{1}{2}\times \left | 6 +20 -4 +9 + 15+2\right |  = 48

The area of the triangle = 48 unit².

5 0
3 years ago
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