how do yu find the implicit differentiation for this porblem
4cosx*siny=1
To find the implicit differentiation of the equation 4cos(x)sin(y) = 1, follow these steps:
Step 1: Recall the differentiation rules for the product, chain, and trigonometric functions.
Step 2: Differentiate both sides of the equation with respect to x, treating y as an implicitly defined function of x.
Step 3: Apply the product rule on the left side of the equation. The product rule states that if u(x) and v(x) are functions of x, then d/dx (u(x)v(x)) = u'(x)v(x) + u(x)v'(x). In this case, let u(x) = 4cos(x) and v(x) = sin(y).
Step 4: Differentiate u(x) = 4cos(x) with respect to x using the chain rule. The chain rule states that if u(g(x)) is a composite function, then d/dx u(g(x)) = u'(g(x))g'(x). In this case, differentiate cos(x) with respect to x to get -sin(x), and since it is multiplied by 4, multiply the result by 4 as well.
Step 5: Differentiate v(x) = sin(y) with respect to x using the chain rule again. The chain rule here states that if y is a function of x, then d/dx sin(y) = cos(y) * y'. Since y is an implicitly defined function of x, we differentiate sin(y) with respect to y to get cos(y) and multiply it by y' (the derivative of y with respect to x).
Step 6: Assemble the parts from step 3, step 4, and step 5 to find the derivative of the left side of the equation.
Step 7: Simplify the derivative expression obtained in step 6.
Let me know if you would like to see the step-by-step calculations for the implicit differentiation of 4cos(x)sin(y) = 1.