Find the derivative of y with respect to x:
y=3sin^4(2-x)^-1
y=[3sin(2-x)^-1]^4
y'=4[3sin(2-x)^-1]^3 (-3cos(2-x)^-1)(-1) -(2-x)^-2
y'=[12cos(2-x)^-1][3sin^3(2-x)^-1][2-x]^2
but the answer does not have a 3 in front of sin. What happened to the 3?
In the step-by-step process you provided, the power rule was correctly applied when differentiating the function y = [3sin(2-x)^-1]^4. However, there seems to be an error in the simplification of the derivative.
Let's go through the differentiation process again step by step:
1. Start with the function y = 3sin^4(2-x)^-1.
2. Apply the chain rule to differentiate the function with respect to x:
y' = 4[3sin(2-x)^-1]^3 (cos(2-x)^-1)(-1)(-1) - (2-x)^-2
3. Simplify the derivative:
y' = 4[3sin(2-x)^-1]^3 (-cos(2-x))/(2-x)^2
Thus, the correct derivative of y with respect to x is:
y' = 4[3sin(2-x)^-1]^3 (-cos(2-x))/(2-x)^2
To find the derivative of the function y=3sin^4(2-x)^-1, let's break down the steps:
Step 1: Rewrite the function
We can rewrite the function as y = (3(sin(2-x)^-1))^4.
Step 2: Apply the chain rule
The derivative of (3(sin(2-x)^-1))^4 can be found using the chain rule.
The chain rule states that if we have a composite function f(g(x)), the derivative is given by f'(g(x)) * g'(x).
In our case, f(u) = u^4 and g(x) = 3(sin(2-x)^-1). Let's find the derivatives of f(u) and g(x) separately.
Derivative of f(u):
Using the power rule, the derivative of u^4 is 4u^3.
Derivative of g(x):
To find the derivative of g(x), we need to use the chain rule again because g(x) involves another function inside it.
Let's let h(x) = sin(2-x), then g(x) = 3 * h(x)^-1.
Derivative of h(x):
Using the chain rule, the derivative of sin(2-x) is -cos(2-x) * (-1) = cos(2-x).
Derivative of g(x):
Applying the chain rule, the derivative of g(x) = 3 * (-1/h(x)^2) * (dh(x)/dx) = -3cos(2-x) * (-1) * h(x)^-2.
Now, we can find the derivative of the function y using the chain rule.
y' = f'(g(x)) * g'(x)
= 4(3(sin(2-x)^-1))^3 * (-3cos(2-x)) * (-1) * (3(sin(2-x)^-1))^-2
Simplifying further:
= 4[3sin^3(2-x)^-1 * (-3cos(2-x) * (3(sin(2-x)^-1))^-2
So, the derivative of y is given by:
y' = 4[3sin^3(2-x)^-1][-3cos(2-x)/(3sin(2-x))^2]