Derivative of (x^4)÷(y^8) with respect to x, given dy÷dx=v
$$\\\frac{d}{dx}\;\frac{x^4}{y^8}\\\\
=\frac{d}{dx}\;x^4*y^{-8}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}\;\;\frac{dy}{dx}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}*v\\\\
=\frac{4x^3}{y^{8}}+\frac{-8x^4v}{y^{9}}\\\\
=\frac{4x^3y-8x^4v}{y^{9}}\\\\
=\frac{4x^3(y-2xv)}{y^{9}}\\\\$$
Could a mathematician check this please ?
Derivative of (x^4)÷(y^8) with respect to x, given dy÷dx=v
$$\\\frac{d}{dx}\;\frac{x^4}{y^8}\\\\
=\frac{d}{dx}\;x^4*y^{-8}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}\;\;\frac{dy}{dx}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}*v\\\\
=\frac{4x^3}{y^{8}}+\frac{-8x^4v}{y^{9}}\\\\
=\frac{4x^3y-8x^4v}{y^{9}}\\\\
=\frac{4x^3(y-2xv)}{y^{9}}\\\\$$
Could a mathematician check this please ?