Differentiation question

Ok so the first part of the question is x = 3sin y find DX/DY in terms of Y
I got 3cos y which is right, then it asked for me to find DY/DX in terms of Y, i then got 1/3 Sec Y, the next part is hence show DY/DX = 1/Root 9 - x to the power of 2, I cant do this part I dont know how to get from 1/3 Sec y to that?

Reply 1

joostan

Original post by SunDun111

Hint:

\cos^2(y)=1-\sin^2(y)

Reply 2

SunDun111

Original post by joostan

Hint:

\cos^2(y)=1-\sin^2(y)

Ey but, isn't it asking me to get from 1/3 Sec y to 1/ Root 9 - x to the power of 2 ,

Reply 3

joostan

Original post by SunDun111

Ey but, isn't it asking me to get from 1/3 Sec y to 1/ Root 9 - x to the power of 2 ,

Yes, but

\sec(y)=\dfrac{1}{\cos(y)}

. . .
Also worth noting that

a\sqrt{f(x)}=\sqrt{a^2f(x)}

Reply 4

SunDun111

Original post by joostan

Yes, but

\sec(y)=\dfrac{1}{\cos(y)}

. . .

I understand, but I'm asking what is the question asking of me? Is it asking me to rearrrange DY/DX in terms of Y to in terms of X?

Reply 5

Zacken

Original post by SunDun111

...

The question wants you to go from

\frac{dy}{dx} = \frac{1}{3\cos y}

into dy/dx as a function of x using the fact that

x = 3\sin y

(edited 8 years ago)

Reply 6

SunDun111

Original post by Zacken

The question wants you to go from

\frac{dy}{dx} = \frac{1}{3\sec y}

into dy/dx as a function of x using the fact that

x = 3\sin y

Right thanks!

Reply 7

Zacken

Original post by SunDun111

Right thanks!

Do you now understand Joostan's (excellent) advice and can apply it here?

(edited 8 years ago)

Reply 8

SunDun111

Original post by Zacken

Do you know understand Joostan's (excellent) advice and can apply it here?

First time i've heard it but trying to,

Reply 9

Zacken

Original post by SunDun111

First time i've heard it but trying to,

Got it?

Reply 10

SunDun111

Original post by Zacken

Got it?

Unforunatley not, i understand the concept and have done other questions like this just now, but this particular one I am struggling to re-arrange, I will ask my teacher when I go back on Monday

Reply 11

Zacken

Original post by SunDun111

Unforunatley not, i understand the concept and have done other questions like this just now, but this particular one I am struggling to re-arrange, I will ask my teacher when I go back on Monday

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{1}{3\cos y} = \frac{1}{3\sqrt{1 - \sin^2 y}} \end{equation*}

since

\cos^2 y + \sin^2 y = 1 \Rightarrow \cos^2 y = 1 - \sin^2 y \Rightarrow \cos y = \sqrt{1 - \sin^2 y}

Now you know that

x = 3\sin y

\frac{x}{3} = \sin y

and

\sin^2 y = \frac{x^2}{9}

.

Plug this back into the centered equation and remember that

3\sqrt{1 -\sin^2 y} \equiv \sqrt{9 - 9\sin^2 y}

Reply 12

SunDun111

Original post by Zacken

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{1}{3\cos y} = \frac{1}{3\sqrt{1 - \sin^2 y}} \end{equation*}

since

\cos^2 y + \sin^2 y = 1 \Rightarrow \cos^2 y = 1 - \sin^2 y \Rightarrow \cos y = \sqrt{1 - \sin^2 y}

Now you know that

x = 3\sin y

\frac{x}{3} = \sin y

and

\sin^2 y = \frac{x^2}{9}

.

Plug this back into the centered equation and remember that

3\sqrt{1 -\sin^2 y} \equiv \sqrt{9 - 9\sin^2 y}

Thanks I got it right, quick question
i am solving the equation 1 - 2ln X I took the 2lnx to the other side, and E'd both sides so i got
E to the power of 1 = 2x
Is the answer e/2?

Reply 13

Zacken

Original post by SunDun111

Thanks I got it right, quick question
i am solving the equation 1 - 2ln X I took the 2lnx to the other side, and E'd both sides so i got
E to the power of 1 = 2x
Is the answer e/2?

No you didn't.

Unparseable latex formula:

\displaystyle [br]\begin{equation*}1 - 2\ln x = 0 \Rightarrow 1= 2\ln x \Rightarrow e^1 = e^{2\ln x} \Rightarrow e = e^{\ln x^2} \Rightarrow x^2 = e \Rightarrow \cdots \end{equation*}

It is not the case that

e^{a \ln f(x)} = a f(x)

.

It is the case that

e^{a\ln f(x)} = e^{\ln f(x)^a} = f(x)^a

(edited 8 years ago)

Reply 14

SunDun111

Original post by Zacken

No you didn't.

Unparseable latex formula:

\displaystyle [br]\begin{equation*}1 - 2\ln x = 0 \Rightarrow 1= 2\ln x \Rightarrow e^1 = e^{2\ln x} \Rightarrow e = e^{\ln x^2} \Rightarrow x^2 = e \Rightarrow \cdots \end{equation*}

It is not the case that

e^{a \ln f(x)} = a f(x)

it is the case that

e^{a\ln f(x)} = e^{\ln f(x)^a} = f(x)^a

I see where I've gone wrong thanks

Reply 15

Zacken

Original post by SunDun111

I see where I've gone wrong thanks

No problem.

Reply 16

SunDun111

Original post by Zacken

No problem.

Sorry for all the questions, doing a lot of revision today

This question is the curve has the equation (x to the power of 2 - 4x +1)e to the minus x, Dy/DX it and find the points,

Dy/dx should be minus e to the minus x (x to the power of 2 - 4x +1) + e to the minus x (2x - 2) = 0 how would I solve this?