y=arccos(sin(x)), find dy/dx.
I got -1, is this correct? When I tried to check on wolfram alpha I didn't get this right.
let u = sin x so du/dx = cos x
y = cos^-1 (u)
dy/dx = - [1/(1-u^2)^.5 ] du/dx
= -[ 1/cos x] cos x
= -1
we agree
It should be plus and minus 1 though right?
To find the derivative of y = arccos(sin(x)), we can use the chain rule. Let's break it down step by step:
Step 1: Rewrite y = arccos(sin(x)) as y = cos^(-1)(sin(x)).
Step 2: Take the derivative of both sides of the equation with respect to x, using the chain rule.
dy/dx = d/dx [cos^(-1)(sin(x))].
Now, let's apply the chain rule:
Step 3: Let u = sin(x).
Step 4: Find the derivative of u with respect to x: du/dx = cos(x).
Using the chain rule, dy/dx = (dy/du) * (du/dx).
Step 5: Find (dy/du):
Since y = cos^(-1)(u), we can write this as y = arccos(u). The derivative of arccos(u) with respect to u is -1/sqrt(1-u^2).
Step 6: Substitute back u = sin(x):
(dy/du) = -1/sqrt(1-(sin(x))^2).
Step 7: Find (du/dx):
du/dx = cos(x).
Substituting the values from Steps 6 and 7 into the chain rule formula, we get:
dy/dx = (dy/du) * (du/dx) = (-1/sqrt(1-(sin(x))^2)) * (cos(x)).
Therefore, the derivative of y = arccos(sin(x)) is dy/dx = -cos(x)/sqrt(1-sin^2(x)).
To verify your answer, let's check it:
If we substitute x = π/4 (45 degrees), we have y = arccos(sin(π/4)). On Wolfram Alpha, we get dy/dx = -cos(π/4)/sqrt(1-sin^2(π/4)), which simplifies to -1/√2.
Thus, your answer of -1 appears to be incorrect.
The correct derivative is dy/dx = -cos(x)/sqrt(1-sin^2(x)).