What is the derivative of \sqrt[4]{x^3+5x^2+x+6} ?

The derivative of \sqrt[4]{x^3+5x^2+x+6} is (3x^2+10x+1)/(4(x^3+5x^2+x+6)^{3/4)}

What is the first derivative of \sqrt[4]{x^3+5x^2+x+6} ?

The first derivative of \sqrt[4]{x^3+5x^2+x+6} is (3x^2+10x+1)/(4(x^3+5x^2+x+6)^{3/4)}

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\frac{3 x ^{2} + 10 x + 1}{4 ( x ^{3} + 5 x ^{2} + x + 6 ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( x ^{3} + 5 x ^{2} + x + 6 ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{3} + 5 x^{2} + x + 6)

= \frac{1}{4 ( x ^{3} + 5 x ^{2} + x + 6 ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{3} + 5 x^{2} + x + 6)

\frac{d}{d x} (x^{3} + 5 x^{2} + x + 6) = 3 x^{2} + 10 x + 1

= \frac{1}{4 ( x ^{3} + 5 x ^{2} + x + 6 ) ^{\frac{3}{4}}} (3 x^{2} + 10 x + 1)

Simplify

\frac{1}{4 ( x ^{3} + 5 x ^{2} + x + 6 ) ^{\frac{3}{4}}} (3 x^{2} + 10 x + 1) : \frac{3 x ^{2} + 10 x + 1}{4 ( x ^{3} + 5 x ^{2} + x + 6 ) ^{\frac{3}{4}}}

= \frac{3 x ^{2} + 10 x + 1}{4 ( x ^{3} + 5 x ^{2} + x + 6 ) ^{\frac{3}{4}}}

\frac{d y}{d t} - 2 t y = - 9 t^{2} e^{t^{2}}

\frac{d y}{d x} = 2 y^{2} + x y^{2}

\int \frac{1}{x ^{2} 2 + 9 x ^{2}} d x

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

5 t^{- \frac{3}{5}}

What is the derivative of \sqrt[4]{x^3+5x^2+x+6} ?
The derivative of \sqrt[4]{x^3+5x^2+x+6} is (3x^2+10x+1)/(4(x^3+5x^2+x+6)^{3/4)}
What is the first derivative of \sqrt[4]{x^3+5x^2+x+6} ?
The first derivative of \sqrt[4]{x^3+5x^2+x+6} is (3x^2+10x+1)/(4(x^3+5x^2+x+6)^{3/4)}