# What is the derivative of #x sin y + y cos x = 1#?

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

To find the derivative of the given implicit equation (x \sin(y) + y \cos(x) = 1), you'll need to use implicit differentiation.

Differentiating each term with respect to (x), you get:

[ \frac{d}{dx}(x \sin(y)) + \frac{d}{dx}(y \cos(x)) = \frac{d}{dx}(1) ]

Apply the product rule for differentiation on each term:

[ \sin(y) + x \cos(y) \frac{dy}{dx} - y \sin(x) - \sin(x) \frac{dx}{dx} = 0 ]

Simplify and solve for (\frac{dy}{dx}):

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

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

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

- How do you find the second derivative of #x^2+y^2=1#?
- If #F(x)=f(xf(xf(x)))# where f(1)=2, f(2)=3, f'(1)=4, f'(2)=5, and f'(3)=6, how do you find F'(1)?
- How do you differentiate #f(x)= (2x+1)(3x+1)^4 # using the product rule?
- What is the derivative of #sqrtx^2#?
- What is the Quotient Rule for derivatives?

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