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3 years ago

How do you find the derivative of y=ln(cos(x))y=ln(cos(x)) ?

1 answer:

7 0

$y=ln(cosx)\\\\We\ know:\\(lnx)'=\frac{1}{x}\ and\ (cosx)'=-sinx\ and\ \{f[g(x)]\}'=f'[g(x)]\cdot g'(x)\\\\therefore:\\\\y'=\frac{1}{cosx}\cdot(-sinx)=-\frac{sinx}{cosx}=-tanx$

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Can someone help please

icang [17]

I would say C because you multiply 3 x H then 3 x4 I think this is the right answer but I'm not so sure though Hope it helped sort of - Ashlynn

8 0

3 years ago

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Please hellp

ankoles [38]

Answer:

he needs 3 bags i did this test

Step-by-step explanation:

4 0

2 years ago

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Show that (2xy3 + cos x)dx + (3x2 y2 − sin y)dy = 0 is exact, and find the solution. Find c if y(0) = π

Licemer1 [7]

We're looking for a solution of the form $F(x,y)=C$ . By the chain rule, this solution should have total differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

and the equation is exact if the mixed second-order partial derivatives of $F$ are equal, i.e. $\frac{\partial^2F}{\partial x\partial y}=\dfrac{\partial^2F}{\partial y\partial x}$ .

The given ODE is exact, since

$\dfrac{\partial(2xy^3+\cos x)}{\partial y}=6xy^2$

$\dfrac{\partial(3x^2y^2-\sin y)}{\partial x}=6xy^2$

Then

$\dfrac{\partial F}{\partial x}=2xy^3+\cos x\implies F(x,y)=x^2y^3+\sin x+f(y)$

$\dfrac{\partial F}{\partial y}=3x^2y^2-\sin y=3x^2y^2+\dfrac{\mathrm df}{\mathrm dy}$

$\dfrac{\mathrm df}{\mathrm dy}=-\sin y\implies f(y)=\cos y+C$

$\implies x^2y^3+\sin x+\cos y=C$

With $y(0)=\pi$ , we get

$\cos\pi=C\implies C=-1$

$\implies\boxed{x^2y^3+\sin x+\cos y=-1}$

6 0

3 years ago

Complete the statement. 30 out of 65 students in the class are bus riders. In lowest terms,

ExtremeBDS [4]

Write as a fraction:
30/65

Find the greatest common factor:
5

Divide both sides by 5:
6/13

Final answer 6/13 or six out of thirteen students take the bus.

7 0

3 years ago

you plant 30 african violet seeds and 9 of them sprout use experimental probability to predict how many will sprout if you plant

Ostrovityanka [42]

Given that out of 30 planted seeds only 9 of the m sprouted, then:
Experimental probability =(number of occurrence of event)/(total number of trials made)
=9/30
=3/10
thus the number of seeds that will sprout given that 20 seeds we planted will be:
3/10×20
=6 seeds

8 0

3 years ago