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## Homework Statement

Solve the stationary points of y=-sinx+cosx for domain -pi<x<pi

## Homework Equations

## The Attempt at a Solution

Differentiate: d/dx=cosx+sinx But how do i solve?

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- Thread starter pip_beard
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- #1

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Solve the stationary points of y=-sinx+cosx for domain -pi<x<pi

Differentiate: d/dx=cosx+sinx But how do i solve?

- #2

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cosx+sinx=0

cosx=-sinx

1=-tanx??

cosx=-sinx

1=-tanx??

- #3

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I dont know how they got these??

because surly the x co-ordinate is 0???

- #4

Mark44

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If y = -sinx + cosx, what is dy/dx? Note that it is incorrect to say "d/dx = ..."## Homework Statement

Solve the stationary points of y=-sinx+cosx for domain -pi<x<pi

## Homework Equations

## The Attempt at a Solution

Differentiate: d/dx=cosx+sinx But how do i solve?

If you meant dy/dx = ..., you have made a mistake. Try again.

Also, your notation is not correct. d/dx is an operator that is written to the left of some function. In contrast, dy/dx is the derivative of y with respect to x.

- #5

Mark44

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Why would you think this?because surly the x co-ordinate is 0???

- #6

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dy/dx=cosx+sinx.

stationary points when diff = 0

so cosx+sinx=0 where do i go from here??

- #7

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Why would you think this?

because stationary points lie on the x axis??

- #8

Mark44

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x-values lie on the x-axis, but stationary points lie on the curve, which might not even touch the x-axis. For example, the only stationary point on the graph of y = x^2 + 1 is at (0, 1). This is not a point on the x-axis.because stationary points lie on the x axis??

- #9

Mark44

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No, dy/dx = -cosx - sinxso therefore:

dy/dx=cosx+sinx.

To find the stationary points, set dy/dx to zero.

-cosx - sinx = 0

==> cosx + sinx = 0

==> 1 + tanx = 0 (dividing both sides by cosx)

Can you continue?

stationary points when diff = 0

so cosx+sinx=0 where do i go from here??

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