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3 years ago

Use logarithmic differentiation to find dy/dx

Mathematics

1 answer:

liq [111]3 years ago

7 0

Answer:

dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

Step-by-step explanation:

y = (x^2 - 3)^sinx

ln y = ln (x^2 - 3)^sinx

ln y = sin x * ln (x^2 - 3)

1/y * dy/dx = sin x * {1 / (x^2 - 3)} * 2x + ln(x^2 - 3) * cos x

1/y dy/dx = 2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)

dy/dx = [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)] * y

dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

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A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

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Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

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Negative sign denotes the distance between the car and the person decrease.

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