What is (\partial)/(\partial x)(ln(tan(x+y))) ?

The answer to (\partial)/(\partial x)(ln(tan(x+y))) is sec^2(x+y)cot(x+y)

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(\partial)/(\partial x)(ln(tan(x+y)))

\frac{\partial}{\partial x} (ln (tan (x + y)))

sec^{2} (x + y) cot (x + y)

Solution steps

\frac{\partial}{\partial x} (ln (tan (x + y)))

Treat

y

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (tan (x + y)) = sec^{2} (x + y)

= \frac{1}{tan ( x + y )} sec^{2} (x + y)

Simplify

\frac{1}{tan ( x + y )} sec^{2} (x + y) : sec^{2} (x + y) cot (x + y)

= sec^{2} (x + y) cot (x + y)

\int_{1}^{- 1} (3 e^{x}) d x

\frac{d}{d x} (x e^{x^{3}})

\frac{1}{( s + 3 ) ^{2} + 4}

x^{2} y^{^{'}} - 2 x^{3} e^{- \frac{1}{x}} = y

\int \frac{sin ( x )}{cos ( 2 x )} d x

What is (\partial)/(\partial x)(ln(tan(x+y))) ?
The answer to (\partial)/(\partial x)(ln(tan(x+y))) is sec^2(x+y)cot(x+y)