find the dy/dx of:
y=3(cscx)^3/2
First you do powers.
3(csc(x))^1/2
Then trig.
The final:
-3(csc(x))^1/2 *(csc(x)cot(x))
thank u for helping
To find the derivative of the given function, you can use the chain rule. Let's go step by step:
Step 1: Rewrite the function using trigonometric identities.
y = 3(csc(x))^3/2
Recall that csc(x) is the reciprocal of sin(x):
csc(x) = 1/sin(x)
Now, substitute this expression back into the original function:
y = 3(1/sin(x))^(3/2)
Step 2: Take the derivative using the chain rule.
Let u = 1/sin(x)
Differentiate u with respect to x:
du/dx = d(1/sin(x))/dx
To find the derivative of u, apply the quotient rule:
du/dx = (sin(x)(0) - 1(cos(x)))/(sin(x))^2
= -cos(x)/(sin(x))^2
= -cos(x)/sin^2(x)
Step 3: Rewrite the original function in terms of u.
y = 3u^(3/2)
Step 4: Apply the chain rule to find dy/dx.
dy/dx = d(3u^(3/2))/dx
= 3 * d(u^(3/2))/du * du/dx
Now, differentiate u^(3/2) with respect to u:
d(u^(3/2))/du = (3/2)u^(1/2)
Finally, multiply this result by du/dx to get the derivative in terms of x:
dy/dx = 3 * (3/2)u^(1/2) * (-cos(x)/sin^2(x))
= -9u^(1/2)*cos(x)/(2sin^2(x))
Step 5: Replace u with the original expression 1/sin(x):
dy/dx = -9(1/sin(x))^(1/2)*cos(x)/(2sin^2(x))
Therefore, the derivative of y = 3(csc(x))^3/2 is:
dy/dx = -9(cos(x))/(2sin^(3/2)(x))