y=arccos(sin(x)), find dy/dx and sketch it's graph(I guess I can do this on wolframalpha after I'm done solving the question). And by arcsin I mean inverse of the expression(written like cos^-1(sin(x)), but is not 1/cos(sinx) but the inverse of cos).
I am not getting this inverse stuff very well, please give a step by step solution. Thanks in advance!!
whenever you have problems like this, draw a right triangle. In this case, you want a triangle with one angle x. If the sides are a,b,c with b opposite angle x, and c the hypotenuse, then
sin(x) = b/c
Now, the angle whose cosine is b/c is the other acute angle of the triangle, namely (π/2 - x).
Okay, what you say makes sense, but what do I do with the (Pi/2)?
To find the derivative of y = arccos(sin(x)), we can use the chain rule. The chain rule states that if we have a function g(f(x)), its derivative with respect to x is given by g'(f(x)) * f'(x).
Step 1:
Let's find the derivative of the inner function, sin(x).
dy/dx = d(arccos(sin(x)))/d(sin(x)) * d(sin(x))/dx
Step 2:
Finding d(arccos(sin(x)))/d(sin(x)):
We know that cos(arccos(x)) = x (by definition of arccos), so cos(arccos(sin(x))) = sin(x).
Differentiating both sides with respect to sin(x):
cos(arccos(sin(x))) = sin(x)
-d(arccos(sin(x)))/d(sin(x)) * sin'(x) = cos(x)
-d(arccos(sin(x)))/d(sin(x)) = cos(x)
So, d(arccos(sin(x)))/d(sin(x)) = cos(x)
Step 3:
Finding d(sin(x))/dx:
Differentiating sin(x) with respect to x gives us:
dy/dx = cos(x) * cos(x) = cos^2(x)
So, dy/dx = cos^2(x)
Now, you can use WolframAlpha or any other graphing tool to sketch the graph of dy/dx = cos^2(x).