What is the d/(dy)(tan(x^2+y)) ?

The d/(dy)(tan(x^2+y)) is sec^2(x^2+y)

What is the first d/(dy)(tan(x^2+y)) ?

The first d/(dy)(tan(x^2+y)) is sec^2(x^2+y)

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\frac{d}{d y} (tan (x^{2} + y))

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

Solution steps

\frac{d}{d y} (tan (x^{2} + y))

Treat

x

as a constant

Apply the chain rule:

sec^{2} (x^{2} + y) \frac{d}{d y} (x^{2} + y)

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

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

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

Simplify

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

x^{3} - 4

\int e^{2 x y} d x

\int cos (2 x) cos (x) d x

\int \frac{x}{( x ^{4} + 1 )} d x

What is the d/(dy)(tan(x^2+y)) ?
The d/(dy)(tan(x^2+y)) is sec^2(x^2+y)
What is the first d/(dy)(tan(x^2+y)) ?
The first d/(dy)(tan(x^2+y)) is sec^2(x^2+y)