Find dy/dx sin(y)+cos(x)=1

$\sin (y) + \cos (x) = 1$

Differentiate both sides of the equation.

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

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $\sin (y) + \cos (x)$ with respect to $x$ is $\frac{d}{d x} [\sin (y)] + \frac{d}{d x} [\cos (x)]$ .

$\frac{d}{d x} [\sin (y)] + \frac{d}{d x} [\cos (x)]$

Evaluate $\frac{d}{d x} [\sin (y)]$ .

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = y$ .

To apply the Chain Rule, set $u$ as $y$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [y] + \frac{d}{d x} [\cos (x)]$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d x} [y] + \frac{d}{d x} [\cos (x)]$

Replace all occurrences of $u$ with $y$ .

$\cos (y) \frac{d}{d x} [y] + \frac{d}{d x} [\cos (x)]$

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\cos (y) y' + \frac{d}{d x} [\cos (x)]$

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$\cos (y) y' - \sin (x)$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0$

Reform the equation by setting the left side equal to the right side.

$\cos (y) y' - \sin (x) = 0$

Solve for $y'$ .

Simplify the left side.

Reorder factors in $\cos (y) y' - \sin (x)$ .

$y' \cos (y) - \sin (x) = 0$

Add $\sin (x)$ to both sides of the equation.

$y' \cos (y) = \sin (x)$

Divide each term in $y' \cos (y) = \sin (x)$ by $\cos (y)$ and simplify.

Divide each term in $y' \cos (y) = \sin (x)$ by $\cos (y)$ .

$\frac{y' \cos (y)}{\cos (y)} = \frac{\sin (x)}{\cos (y)}$

Simplify the left side.

Cancel the common factor of $\cos (y)$ .

Cancel the common factor.

$\frac{y' \cos (y)}{\cos (y)} = \frac{\sin (x)}{\cos (y)}$

Divide $y'$ by $1$ .

$y' = \frac{\sin (x)}{\cos (y)}$

Simplify the right side.

Rewrite $\frac{\sin (x)}{\cos (y)}$ as a product.

$y' = \sin (x) \frac{1}{\cos (y)}$

Write $\sin (x)$ as a fraction with denominator $1$ .

$y' = \frac{\sin (x)}{1} \cdot \frac{1}{\cos (y)}$

Simplify.

Divide $\sin (x)$ by $1$ .

$y' = \sin (x) \frac{1}{\cos (y)}$

Convert from $\frac{1}{\cos (y)}$ to $\sec (y)$ .

$y' = \sin (x) \sec (y)$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \sin (x) \sec (y)$

Find dy/dx sin(y)+cos(x)=1

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