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2 years ago

Find the derivative of

Mathematics

2 answers:

Anit [1.1K]2 years ago

6 0

Answer:

⇒ $y'=-\frac{2}{(six-cosx)^{2} }$

Step-by-step explanation:

⇒Answer in the attachment.

Gre4nikov [31]2 years ago

3 0

Answer:

<u><em>y' = 1 / sinxcosx</em></u>

Step-by-step explanation:

<u>Given :</u>

y = sinx + cosx / sinx - cosx

<u>Applying quotient rule :</u>

y' = [(sinx - cosx)(cosx - sinx) - (sinx + cosx)(cosx + sinx)] / (sinx -

cosx)²

y' = [sinxcosx - cos²x - sin²x + sinxcosx - (sinxcosx + cos²x + sin²x + sinxcosx)] / (sinx - cosx)²

y' = 2sinxcosx - 2sinxcosx - 2cos²x - 2sin²x / (sinx -cosx)²

y' = -2(sin²x + cos²x) / (sinx - cosx)²

y' = -2 / (sinx - cosx)²

y' = -2 / sin²x - 2sinxcosx + cos²x

y' = -2 / -2sinxcosx

<u><em>y' = 1 / sinxcosx</em></u> [⇒ Final answer]

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The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

Hitman42 [59]

Answer:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Step-by-step explanation:

The probability (P) to find the particle is given by:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

The solution of the intregral of equation (1) is:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

I hope it helps you!

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4 years ago

John is creating a Thanksgiving display at the store where he works, using only canned pumpkin and canned green beans. He needs

nadya68 [22]

Well 3x(pumpkin) + 1x ( beans ) =4x so 4x=76
divide 76 by four to get x=19
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3 years ago

A stadium floor that is in the shape of a circle has a diameter with a length of 50 yards. What is the area of the circle on the

barxatty [35]

Step-by-step explanation:

as diameter=50 yards

radius (r)=50/2=25 yards

ar of circle = π r r

π=22/7

22÷7×25×25

1974.285714

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3 years ago

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Find f(t – 3) for f(x) = 4x^2 – 8x + 4.

Vladimir79 [104]

Answer:

Step-by-step explanation:

Instead of x put (t - 3) in the equation of the function f(x) = 4x² - 8x + 4:

f(t - 3) = 4(t - 3)² - 8(t - 3) + 4

<em>use (a - b)² = a² - 2ab + b² and the distributive property a(b + c) = ab + ac</em>

f(t - 3) = 4(t² - (2)(t)(3) + 3²) + (-8)(t) + (-8)(-3) + 4

f(t - 3) = 4(t² - 6t + 9) - 8t + 24 + 4

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3 years ago

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kotykmax [81]

Answer:

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3 years ago

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