Derivative of (x^1/2 cscX sinX)
The way you typed it ....
(x^1/2 cscX sinX)
= √x (1/sinx)(sinx)
= x^(1/2)
if y = x^(1/2)
dy/dx = (1/2)x^(-1/2) or 1/(2√x)
Not sure that's what you meant, why is the x in x^(1/2) different
than the x in cscX and sinX ?
just use the product rule
y = √x cscx sinx
actually, since cscx sinx = 1,
y' = 1/2 x^-1/2
ah, Ryan, we seem to be doing a lot of very similar problems for you. It is time to try and post your attempts.
To find the derivative of the function f(x) = x^(1/2) csc(x) sin(x), we can use the product rule and chain rule.
The product rule states that if we have two functions u(x) and v(x), the derivative of their product f(x) = u(x) * v(x) is given by:
f'(x) = u'(x) * v(x) + u(x) * v'(x)
In this case, u(x) = x^(1/2) and v(x) = csc(x) * sin(x).
Using the chain rule, we can find the derivatives of u(x) and v(x):
u'(x) = (1/2) * x^(-1/2)
v'(x) = (csc(x) * sin(x))' = csc(x) * sin(x) * cot(x) + csc(x) * cos(x)
Now we can apply the product rule:
f'(x) = u'(x) * v(x) + u(x) * v'(x)
= ((1/2) * x^(-1/2)) * (csc(x) * sin(x)) + (x^(1/2)) * (csc(x) * sin(x) * cot(x) + csc(x) * cos(x))
Simplifying further:
f'(x) = (1/2) * x^(-1/2) * csc(x) * sin(x) + (1/2) * x^(-1/2) * csc(x) * sin(x) * cot(x) + (1/2) * x^(-1/2) * csc(x) * cos(x)
So, the derivative of f(x) = x^(1/2) csc(x) sin(x) is:
f'(x) = (1/2) * x^(-1/2) * csc(x) * sin(x) + (1/2) * x^(-1/2) * csc(x) * sin(x) * cot(x) + (1/2) * x^(-1/2) * csc(x) * cos(x)