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3 years ago

Find the derivative dy / dx for y = x^3 - 5x / x^2 - 1

1 answer:

soldier1979 [14.2K]3 years ago

8 0

$Use\\\\\left(\dfrac{h(x)}{g(x)}\right)'=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}\\\\\text{We have}\\\\h(x)=x^3-5x\to h'(x)=3x^2-5\\\\g(x)=x^2-1\to g'(x)=2x\\\\f(x)=\dfrac{x^3-5x}{x^2-1}\to f'(x)=\left(\dfrac{x^3-5x}{x^2-1}\right)'\\\\=\dfrac{(x^3-5x)'(x^2-1)-(x^3-5x)(x^2-1)'}{(x^2-1)^2}\\\\=\dfrac{(3x^2-5)(x^2-1)-(x^3-5x)(2x)}{(3x^2-5)^2}\\\\=\dfrac{3x^4-3x^2-5x^2+5-2x^4+10x^2}{(x^2-1)^2}\\\\=\dfrac{x^4+2x^2+5}{(x^2-1)^2}\\\\Answer:\ \left(\dfrac{x^3-5x}{x^2-1}\right)'=\dfrac{x^4+2x^2+5}{(x^2-1)^2}$

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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

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What is the sign of y divided by x

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A cylinder has a surface area of 502.4 square inches and a height of 11 inches. What is the volume? PLEASE HELP!!

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Step-by-step explanation:

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3 years ago

Which description best compares the graphs given by the equations:

nexus9112 [7]

x - 5y = -5, -5x - 25y = 25

First, you'll need to get the x variable by itself.

x - 5y = -5
 +5 +5
 x = 0
So x is plotted on the 0.

For the second part of the first equation, you'll be looking for what the y variable represents.

x - 5y = -5
-x -x
 -5y = -5
 5 5
 y = 1
So y is plotted on the 1 on the vertical line above the 0.

For the first part of the second equation, you'll do the same thing as in the first equation.

-5x - 25y = 25
 +25 +25
 -5x = 50
 5 5
x = 10
So the x for this equation is plotted on 10 on the horizontal line.

For the second part of the second equation, you will do the same thing as in the first equation.

-5x - 25y = 25
+5 +5
 -25y = 30
 25 25
y = 1.2
So the y for the second half of the second question is plotted on 1.2 on the vertical line.

<h2>Answer: B) Perpendicular</h2>

Perpendicular means the lines may or may not be of equal length and they will not be perfectly in line with each other.

Parallel means the lines may or may not be of equal length but will be perfectly in line with each other.

Intersecting means the lines may or may not be of equal length but will touch each other.

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2 years ago