Differentiation Question

[Chingoo]

Given that y = (1-sinx)/cosx, show that dy/dx = 1/(1+sinx).

It seems simple but idk why I'm not getting the answer =/

[Dibss]

Let S be sinx & C be cosx.
Use the y=u/v => dy/dx = [Udv/dx - Vdu/dx]/V²
Therefore
dy/dx = [(1-S)(-S) - C(0-C)]/C²

[-S + S² + C²]/C²

Using identity: C²=1-S²

[-S + S² + 1 - S²]/(1-S²)

(-S+1)/(1-S²⁾)

Use identity: a^2 - b^2 = (a+b) (a-b)

(1-S)/(1+S)(1-S)

Finally dy/dx = 1/(1+sinx)

(:

[Arthur Bon Zavi]

Let S be sinx & C be cosx.
Use the y=u/v => dy/dx = [Udv/dx - Vdu/dx]/V²
Therefore
dy/dx = [(1-S)(-S) - C(0-C)]/C²

[-S + S² + C²]/C²

Using identity: C²=1-S²

[-S + S² + 1 - S²]/(1-S²)

(-S+1)/(1-S²⁾)

Use identity: a^2 - b^2 = (a+b) (a-b)

(1-S)/(1+S)(1-S)

Finally dy/dx = 1/(1+sinx)

(:

 + rep.

[Dibss]

Thanks. (:

[Chingoo]

Thanks, but now I'm confused. Isn't the quotient rule for differentiation
y=u/v => dy/dx = [v.du/dx - u.dv/dx]/v^2
Granted you managed to get the answer, but I'm pretty sure that's the quotient rule.

[Dibss]

Sorry, I haven't used the rule for months now - rusty. :S
My bad. So you probably got the answer after using the correct rule right?

[Dibss]

&& you get the correct answer using the product rule

y=(1-S)(C^-1)

dy/dx = (1-S)(S/C²)+(C^-1)(0-C)
=S(1-S))/(1-S²) + 1/C (-C)
=S(1-S)/(1+S)(1-S)  - 1
=(S + 1 - S)/(1+S)
=1/(1+S)

[Chingoo]

Thanks a lot! =D +rep

[Dibss]

No problemm. (:

[Tohru Kyo Sohma]

i kind of find maths AS very hard....especially differentiation.....i have a doubt here:
find the coordinates of the point on the curve y=2x^2 -x-1, where the tangent is parallel to the line; y=3x-2?

[Dibss]

Practice and you'll get used to it eventually. (:

gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:

[Arthur Bon Zavi]

Differentiation is easy and Integration is easier.

[Tohru Kyo Sohma]

Practice and you'll get used to it eventually. (:

gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:
thank you so much wispher
i got it now...........

[Tohru Kyo Sohma]

i have a doubt here;
The production cost per kilogram C(in thousand pounds) when x kilogram of a chemical are made, is given by;
C= 3x+100/x , x>0. find the value for x for which the cost is minimum and the minimum cost??

[elemis]

Differentiate the equation. Set dy/dx = 0 and solve for x.

Determine if the stationary point is a minimum or maximum. Select the minimum point and plug it into the original equation.

This is the minimum cost.