Derivative Help

[kggirl]

Please help me with this problem:

Find dy/dx

y = cosx/x + x/cosx

I think I start out with this:

-sin(x)/x + x/-sinx but I don't know what to do after that or if that is the correct step to take

 

[stapel]

I'm not sure how you got what you did. Please reply showing the steps you used when you applied the Quotient Rule to each of the two terms.

Thank you.

Eliz.

 

[kggirl]

I'm not sure how you got what you did. Please reply showing the steps you used when you applied the Quotient Rule to each of the two terms.

Thank you.

Eliz.

cos x = -sinx, so I just replaced the cosx in the equation with -sinx:

y = cosx/x + x/cos x so I got -sinx/x + x/-sinx

 

[stapel]

Does the Quotient Rule say that, for h(x) = f(x)/g(x), the derivative is h'(x) = f'(x)/g(x), or does it say something rather different?

Eliz.

 

[kggirl]

Does the Quotient Rule say that, for h(x) = f(x)/g(x), the derivative is h'(x) = f'(x)/g(x), or does it say something rather different?

Eliz.

Looking at the rule I have it says: d/dx which is u/v =

Vdu/dx - Udv/dx
______________
V^2

Using that I get:

x-sinx-cos x -sinx - cosx
___________ + _________
x^2 cos ^2x

Am I heading in the right direction?

 

[stapel]

If you mean the following:

. . . . .[x - sin(x) - cos(x)] / [x²] + [-sin(x) - cos(x)] / [cos²(x)]

...then you are moving in the right direction. But how did you get three terms for the numerator of your first fraction?

The Quotient Rule says:

. . . . .For f(x)/g(x), the derivative is given by:

. . . . .[f'(x)g(x) - f(x)g'(x)] / [g(x)]²

For the first fraction, cos(x)/x, let f(x) = cos(x) and g(x) = x. Then f'(x) = -sin(x) and g'(x) = 1. Plug into the formula, and simplify.

Eliz.

 

[kggirl]

If you mean the following:

. . . . .[x - sin(x) - cos(x)] / [x²] + [-sin(x) - cos(x)] / [cos²(x)]

...then you are moving in the right direction. But how did you get three terms for the numerator of your first fraction?

The Quotient Rule says:

. . . . .For f(x)/g(x), the derivative is given by:

. . . . .[f'(x)g(x) - f(x)g'(x)] / [g(x)]²

For the first fraction, cos(x)/x, let f(x) = cos(x) and g(x) = x. Then f'(x) = -sin(x) and g'(x) = 1. Plug into the formula, and simplify.

Eliz.

OK-- is this right:
cosx/ x + x/cos x = f(x)/g(x)/g(x)^2
for cos x/x
f(x) = cos x = -sin x
g(x) = x =1
and for x/cosx
f(x) = x = 1
g(x) = cosx = -sinx therefore,

x-sinx-cosx(1)/x^2 + (1)cosx-1(-sinx)/cos(x)2 =
x-sinx-cosx/x^2 + cosx + sin x/cos2(x)

 

[stapel]

Where did you get that the Quotient Rule says f(x)/g(x) + g(x)/g(x) "=" f(x)/g(x)/(g(x))²?

Just use the rule as it was given to you. Try differentiating just the first of the two fractions, using cos(x) as f(x), x as g(x), -sin(x) as f'(x), and 1 as g'(x). Just plug these into the given formula.

Eliz.