What is (\partial)/(\partial x)(cos(x/(1+y))) ?

The answer to (\partial)/(\partial x)(cos(x/(1+y))) is -(sin(x/(1+y)))/(1+y)

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(\partial)/(\partial x)(cos(x/(1+y)))

\frac{\partial}{\partial x} (cos (\frac{x}{1 + y}))

- \frac{sin ( \frac{x}{1 + y} )}{1 + y}

Solution steps

\frac{\partial}{\partial x} (cos (\frac{x}{1 + y}))

Treat

y

as a constant

Apply the chain rule:

- sin (\frac{x}{1 + y}) \frac{\partial}{\partial x} (\frac{x}{1 + y})

= - sin (\frac{x}{1 + y}) \frac{\partial}{\partial x} (\frac{x}{1 + y})

\frac{\partial}{\partial x} (\frac{x}{1 + y}) = \frac{1}{1 + y}

= - sin (\frac{x}{1 + y}) \frac{1}{1 + y}

Simplify

- sin (\frac{x}{1 + y}) \frac{1}{1 + y} : - \frac{sin ( \frac{x}{1 + y} )}{1 + y}

= - \frac{sin ( \frac{x}{1 + y} )}{1 + y}

\int x^{2} - 6 x + 34 d x

x \to 2 + lim (in t x)

\frac{d}{d θ} (sin^{2} (θ))

y^{^{''}} + 49 y = 3 tan (7 x)

f (x) = 2 x + 7, at x = 9

What is (\partial)/(\partial x)(cos(x/(1+y))) ?
The answer to (\partial)/(\partial x)(cos(x/(1+y))) is -(sin(x/(1+y)))/(1+y)