What is (\partial)/(\partial x)(tan(x^3+y^2)) ?

The answer to (\partial)/(\partial x)(tan(x^3+y^2)) is sec^2(x^3+y^2)*3x^2

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(\partial)/(\partial x)(tan(x^3+y^2))

\frac{\partial}{\partial x} (tan (x^{3} + y^{2}))

sec^{2} (x^{3} + y^{2}) \cdot 3 x^{2}

Solution steps

\frac{\partial}{\partial x} (tan (x^{3} + y^{2}))

Treat

y

as a constant

Apply the chain rule:

sec^{2} (x^{3} + y^{2}) \frac{\partial}{\partial x} (x^{3} + y^{2})

= sec^{2} (x^{3} + y^{2}) \frac{\partial}{\partial x} (x^{3} + y^{2})

\frac{\partial}{\partial x} (x^{3} + y^{2}) = 3 x^{2}

= sec^{2} (x^{3} + y^{2}) \cdot 3 x^{2}

\int \frac{2}{2 x ^{2} + 3 x + 1} d x

x \to 3 + lim (\frac{x + 3}{x - 3})

\frac{\partial}{\partial x} (\frac{x}{x ^{6} - y ^{7}})

n = 4 \sum \infty (x + 1)^{n}

f (x) = x^{- 8}

What is (\partial)/(\partial x)(tan(x^3+y^2)) ?
The answer to (\partial)/(\partial x)(tan(x^3+y^2)) is sec^2(x^3+y^2)*3x^2