What is (\partial)/(\partial x)(6-arctan(7x^2+2y^2)) ?

The answer to (\partial)/(\partial x)(6-arctan(7x^2+2y^2)) is -(14x)/((7x^2+2y^2)^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial x)(6-arctan(7x^2+2y^2))

\frac{\partial}{\partial x} (6 - arctan (7 x^{2} + 2 y^{2}))

- \frac{14 x}{( 7 x ^{2} + 2 y ^{2} ) ^{2} + 1}

Solution steps

\frac{\partial}{\partial x} (6 - arctan (7 x^{2} + 2 y^{2}))

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (6) - \frac{\partial}{\partial x} (arctan (7 x^{2} + 2 y^{2}))

\frac{\partial}{\partial x} (6) = 0

\frac{\partial}{\partial x} (arctan (7 x^{2} + 2 y^{2})) = \frac{14 x}{( 7 x ^{2} + 2 y ^{2} ) ^{2} + 1}

= 0 - \frac{14 x}{( 7 x ^{2} + 2 y ^{2} ) ^{2} + 1}

Simplify

= - \frac{14 x}{( 7 x ^{2} + 2 y ^{2} ) ^{2} + 1}

\int t ln (t + 2) d t

\frac{d}{d t} (sin (5 t))

y^{^{'}} = 2 x - 3 y

\int 2 (tan^{2} (x) + tan^{4} (x)) d x

\int x arccot (x) d x

What is (\partial)/(\partial x)(6-arctan(7x^2+2y^2)) ?
The answer to (\partial)/(\partial x)(6-arctan(7x^2+2y^2)) is -(14x)/((7x^2+2y^2)^2+1)