y = s*square root of(1-s^2) + cos inverse(s)
Just give me some hints and I will do it. Thanks.
You have
y = s*square root of(1-s^2) + cos inverse(s)
which I would write as
y = s*sqrt(1-s2) + cos-1(s)
and you want dy/ds
For the first term use the product rule. You should know how to differentiate the square root by now if you're doing a problem like this.
For the second term, cos-1(u)
d/du cos-1(u) = -1/sqrt(1-u2)
Can you take it from here?
Sure! Let's start with finding the derivative of the first term using the product rule.
The product rule states that if you have a function u(x) multiplied by another function v(x), the derivative of the product can be found using the following formula:
d/dx (u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
In your case, the function u(x) is s and the function v(x) is √(1-s^2).
So, let's find the derivatives of each term:
u'(x) = 1 (since s is a constant with respect to s)
v'(x) = d/dx (√(1-s^2))
To find the derivative of the square root function, we can use the chain rule. The chain rule states that if you have a composite function g(f(x)), the derivative can be found using the following formula:
d/dx (g(f(x))) = g'(f(x)) * f'(x)
In this case, g(x) = √x and f(x) = 1-s^2.
The derivative of g(x) = √x is 1/(2√x).
The derivative of f(x) = 1-s^2 is -2s.
Now, let's put it all together:
v'(x) = d/dx (√(1-s^2)) = (1/(2√(1-s^2))) * (-2s)
Now, we can substitute u'(x) and v'(x) back into the product rule formula:
d/ds (s*√(1-s^2)) = s * (-2s/(2√(1-s^2))) + 1 * √(1-s^2)
Simplifying this expression gives us:
d/ds (s*√(1-s^2)) = -s^2/√(1-s^2) + √(1-s^2)
For the second term, we have cos^(-1)(s). To find its derivative, we can use the chain rule again:
d/ds (cos^(-1)(s)) = (-1)/√(1-s^2)
Now that we have both derivatives, we can add them together to get the final result:
dy/ds = -s^2/√(1-s^2) + √(1-s^2) - (1)/√(1-s^2)
Simplifying this expression further gives us the derivative of y with respect to s.