Using the product rule, find the derivative of the following function:
(x^1/2 cscx sinx).
First off, csc * sin = 1
So, y = √x
However, if you insist on doing it the hard way
y = √x cscx sinx
y' = 1/2√x cscx sinx + √x(cscx sinx)'
= 1/2√x * 1 + √x(-cscx ctnx * sinx + cscx cosx)
= 1/2√x + √x(-ctnx + ctnx)
= 1/2√x
same answer
To find the derivative of the function (x^(1/2) csc(x) sin(x)), we will use the product rule. The product rule states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function.
Let's break down the given function into two separate functions:
f(x) = x^(1/2) and g(x) = csc(x) sin(x)
Now, let's find the derivatives of each function separately:
f'(x) = (1/2)x^(-1/2) (by applying the power rule for differentiation)
g'(x) = (csc(x) cos(x) - cot(x) sin(x)) (by applying the chain rule and the derivative of csc(x))
Now, using the product rule, we can find the derivative of the given function:
(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
Plugging in the values we found earlier:
(f(x)g(x))' = [(1/2)x^(-1/2)](csc(x) sin(x)) + (x^(1/2))(csc(x) cos(x) - cot(x) sin(x))
Simplifying this expression will give you the derivative of the given function.