What is the derivative of \sqrt[4]{2x-3y} ?

The derivative of \sqrt[4]{2x-3y} is (2-3\frac{dy)/(dx)}{4(2x-3y)^{3/4}}

What is the first derivative of \sqrt[4]{2x-3y} ?

The first derivative of \sqrt[4]{2x-3y} is (2-3\frac{dy)/(dx)}{4(2x-3y)^{3/4}}

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\frac{2 - 3 \frac{d y}{d x}}{4 ( 2 x - 3 y ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( 2 x - 3 y ) ^{\frac{3}{4}}} \frac{d}{d x} (2 x - 3 y)

= \frac{1}{4 ( 2 x - 3 y ) ^{\frac{3}{4}}} \frac{d}{d x} (2 x - 3 y)

\frac{d}{d x} (2 x - 3 y) = 2 - 3 \frac{d y}{d x}

= \frac{1}{4 ( 2 x - 3 y ) ^{\frac{3}{4}}} (2 - 3 \frac{d y}{d x})

Simplify

\frac{1}{4 ( 2 x - 3 y ) ^{\frac{3}{4}}} (2 - 3 \frac{d y}{d x}) : \frac{2 - 3 \frac{d y}{d x}}{4 ( 2 x - 3 y ) ^{\frac{3}{4}}}

= \frac{2 - 3 \frac{d y}{d x}}{4 ( 2 x - 3 y ) ^{\frac{3}{4}}}

\int_{\frac{π}{4}}^{π} sin (x) d x

\frac{d}{d x} (cos^{4} (x) + sec (x^{3}))

\int cos (x) d x

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

f (x) = 4 x + 1

What is the derivative of \sqrt[4]{2x-3y} ?
The derivative of \sqrt[4]{2x-3y} is (2-3\frac{dy)/(dx)}{4(2x-3y)^{3/4}}
What is the first derivative of \sqrt[4]{2x-3y} ?
The first derivative of \sqrt[4]{2x-3y} is (2-3\frac{dy)/(dx)}{4(2x-3y)^{3/4}}