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3 years ago

10

If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10

. Show that P must move on a line.

1 answer:

Ksivusya [100]3 years ago

3 0

Answer:

I need points!!!!!!!!

Step-by-step explanation:

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3 years ago

Find dy du , du dx , and dy dx .

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a) $\frac{dy}{du}=5(x^2 +1)^4$ , $\frac{du}{dx}=2x$ , $\frac{dy}{dx}=10x(x^2+1)^4$

b) $\frac{dy}{du}=4(4x^2-x+6)^3$ , $\frac{du}{dx}=8x-1$ , $\frac{dy}{dx}=4(8x-1)(4x^2-2+6)^3$

Step-by-step explanation:

We can use the chain rule in the following form: is u=u(x) is a differentiable function depending on x and y=y(u) is a differentiable function depending on u, then $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$ .

a) $\frac{dy}{du}=\frac{d}{du} (u^5)=5u^4=5(x^2 +1)^4$ from the power rule.

$\frac{du}{dx}=\frac{d}{dx} (x^2 +1)=2x$ .

From the previous parts and the chain rule, $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=5(x^2 +1)^4(2x)=10x(x^2+1)^4$

b) $\frac{dy}{du}=\frac{d}{du} (u^4)=4u^3=4(4x^2-x+6)^3$

$\frac{du}{dx}=\frac{d}{dx} (4x^2-x+6)=8x-1$ from the power and sum rules.

Then, $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=4(4x^2 -x+6)^3(8x-1)=4(8x-1)(4x^2-x+6)^3$

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