# How do you implicitly differentiate # x cos(y) + y cos(x) = 1 ?

By signing up, you agree to our Terms of Service and Privacy Policy

To implicitly differentiate ( x \cos(y) + y \cos(x) = 1 ), we apply the chain rule and product rule. The derivative of ( x \cos(y) ) with respect to ( x ) is ( \cos(y) - x \sin(y) \frac{dy}{dx} ), and the derivative of ( y \cos(x) ) with respect to ( x ) is ( -\sin(x) \frac{dy}{dx} + \cos(x) ). Then, we solve for ( \frac{dy}{dx} ). The resulting derivative is:

[ \frac{dy}{dx} = \frac{\sin(x) - x \sin(y)}{\sin(y) - y \sin(x)} ]

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the derivative of #y=1+x^-1+x^-2+x^-3#?
- How do you implicitly differentiate #-1=-y^2x-2xy+ye^x #?
- How do you find #(dy)/(dx)# given #4x=-5y^2-x^2y+4#?
- How do you differentiate #f(x)=tan(sqrt(x^2-3)) # using the chain rule?
- How do you differentiate #g(x) =(2x^2+1)^(3/2) (x-1)^(7/4) # using the product rule?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7