Which of the following expressions represents y′ when sinx+cosy=sin(x2)+cos(y2) ? (1 point) Responses y′=2xsin(x2)−sinx2ycos(y2)−cosy y ' = 2 x sin ( x 2 ) − sin x 2 y cos ( y 2 ) − cos y y′=2xsin(x2)+sinx2ycos(y2)+cosy y ' = 2 x sin ( x 2 ) + sin x 2 y cos ( y 2 ) + cos y y′=2xcos(x2)−cosx2ysin(y2)−siny y ' = 2 x cos ( x 2 ) − cos x 2 y sin ( y 2 ) − sin y y′=2xcos(x2)+cosx2ysin(y2)+siny y ' = 2 x cos ( x 2 ) + cos x 2 y sin ( y 2 ) + sin y y′=2ycos(y2)−cosy2xsin(x2)−sinx
To find the expression for \( y' \) (which represents the derivative of \( y \) with respect to \( x \), often noted as \( \frac{dy}{dx} \) ), we need to differentiate both sides of the given equation implicitly with respect to \( x \). Our equation is:
\[ \sin(x) + \cos(y) = \sin(x^2) + \cos(y^2) \]
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}[\sin(x) + \cos(y)] = \frac{d}{dx}[\sin(x^2) + \cos(y^2)] \]
Using the chain rule and the fact that \( \frac{dy}{dx}=y' \), we get:
\[ \cos(x) - \sin(y)y' = 2x\cos(x^2) - 2y\sin(y^2)y' \]
Now, solve for \( y' \):
\[ \cos(x) - 2x\cos(x^2) = y'(\sin(y) + 2y\sin(y^2)) \]
\[ y' = \frac{\cos(x) - 2x\cos(x^2)}{\sin(y) + 2y\sin(y^2)} \]
None of the provided choices exactly match the expression derived here. However, the expression for \( y' \) that most correctly represents the differentiation process and simplification of the given equation would be:
\[ y' = \frac{2x\cos(x^2) - \cos(x)}{2ysin(y^2) + \sin(y)} \]
If we look at the given choices and try to match to our derived expression:
- The first and third options incorrectly mix sine and cosine in the numerator.
- The second and fourth options incorrectly have a plus sign between the terms in the numerator.
- The last option flips the \( x \) and \( y \) terms incorrectly.
Therefore, none of the provided options are correct. Note that the cosine and sine functions in the options provided suggest a mix-up or typo that doesn't align with the correct differentiation process. The correct answer should look more like the expression we derived, with appropriate signs and terms as described.