Let #siny + cosx = 1# then find #(d^2y)/(dx^2)#?

Question

Scarlett Allison · Accepted Answer

# (d^2y)/(dx^2) = siny/cosy \ sin^2x/cos^2y + cosx /cosy #

We have: # siny+cosx=1 # If we differentiate (implicitly) wrt #x# we get: # cosy \ dy/dx -sinx = 0 # .... [A] If we differentiate (implicitly) wrt #x# again whilst applying the product rule we get: # (cosy) \ (d^2y)/(dx^2) -(siny \ dy/dx) \ dy/dx - cosx = 0 # # :. (d^2y)/(dx^2) -siny/cosy \ (dy/dx)^2 - cosx/cosy = 0 # And, from [A] we get #dy/dx = sinx/cosy#, so substituting this result in the above we find that: # (d^2y)/(dx^2) -siny/cosy \ (sinx/cosy)^2 - cosx /cosy= 0 # # :. (d^2y)/(dx^2) = siny/cosy \ sin^2x/cos^2y + cosx /cosy #

Isabella Ackerman · Answer

undefined To find the second derivative of $ y $ with respect to $ x $, given $ \sin y + \cos x = 1 $, we need to find $ \frac{{d^2y}}{{dx^2}} $.

1. Start by finding the first derivative of $ y $ with respect to $ x $ using implicit differentiation.

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

$$ \cos y \cdot \frac{{dy}}{{dx}} - \sin x = 0 $$

2. Now, we'll differentiate again to find the second derivative.

$$ \frac{{d}}{{dx}}(\cos y \cdot \frac{{dy}}{{dx}}) - \frac{{d}}{{dx}}(\sin x) = 0 $$

$$ -\sin y \cdot (\frac{{dy}}{{dx}})^2 + \cos y \cdot \frac{{d^2y}}{{dx^2}} - \cos x = 0 $$

3. Since $ \sin y + \cos x = 1 $, we can substitute $ \cos x = 1 - \sin y $ into the equation.

$$ -\sin y \cdot (\frac{{dy}}{{dx}})^2 + \cos y \cdot \frac{{d^2y}}{{dx^2}} - (1 - \sin y) = 0 $$

4. Rearrange and solve for $ \frac{{d^2y}}{{dx^2}} $.

$$ \cos y \cdot \frac{{d^2y}}{{dx^2}} = (1 - \sin y) + \sin y \cdot (\frac{{dy}}{{dx}})^2 $$

$$ \frac{{d^2y}}{{dx^2}} = \frac{{(1 - \sin y) + \sin y \cdot (\frac{{dy}}{{dx}})^2}}{{\cos y}} $$

5. Finally, substitute $ \sin y + \cos x = 1 $ into the equation.

$$ \frac{{d^2y}}{{dx^2}} = \frac{{(1 - \sin y) + \sin y \cdot (\frac{{dy}}{{dx}})^2}}{{\sqrt{1 - (\sin y)^2}}} $$

This is the expression for the second derivative of $ y $ with respect to $ x $.