What is (\partial)/(\partial x)((a^2x^2+b^2y^2)^4) ?

The answer to (\partial)/(\partial x)((a^2x^2+b^2y^2)^4) is 8a^2x(a^2x^2+b^2y^2)^3

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(\partial)/(\partial x)((a^2x^2+b^2y^2)^4)

\frac{\partial}{\partial x} ((a^{2} x^{2} + b^{2} y^{2})^{4})

8 a^{2} x (a^{2} x^{2} + b^{2} y^{2})^{3}

Solution steps

\frac{\partial}{\partial x} ((a^{2} x^{2} + b^{2} y^{2})^{4})

Treat

a, b, y

as constants

Apply the chain rule:

4 (a^{2} x^{2} + b^{2} y^{2})^{3} \frac{\partial}{\partial x} (a^{2} x^{2} + b^{2} y^{2})

= 4 (a^{2} x^{2} + b^{2} y^{2})^{3} \frac{\partial}{\partial x} (a^{2} x^{2} + b^{2} y^{2})

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

= 4 (a^{2} x^{2} + b^{2} y^{2})^{3} \cdot 2 a^{2} x

Simplify

= 8 a^{2} x (a^{2} x^{2} + b^{2} y^{2})^{3}

(ln (2)) x

f (x) = x^{2} - 2 x + 1

\frac{d}{d x} (\frac{x ^{3}}{3} + c)

x \to 5 lim (\frac{x ^{3} + 8}{x + 2})

\frac{d}{d x} (x - \frac{1}{1 + x})

What is (\partial)/(\partial x)((a^2x^2+b^2y^2)^4) ?
The answer to (\partial)/(\partial x)((a^2x^2+b^2y^2)^4) is 8a^2x(a^2x^2+b^2y^2)^3