What is the slope of-10x^3+6x^{40}y+y^4=3 ?

The slope of-10x^3+6x^{40}y+y^4=3 is (15x^2(-8x^{37}y+1))/(3x^{40)+2y^3}

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slope of-10x^3+6x^{40}y+y^4=3

slope

- 10 x^{3} + 6 x^{40} y + y^{4} = 3

\frac{15 x ^{2} ( - 8 x ^{37} y + 1 )}{3 x ^{40} + 2 y ^{3}}

Solution steps

- 10 x^{3} + 6 x^{40} y + y^{4} = 3

Use implicit differentiation to find the slope of

- 10 x^{3} + 6 x^{40} y + y^{4} = 3

Implicit Derivative

\frac{d y}{d x}

- 10 x^{3} + 6 x^{40} y + y^{4} = 3 : \frac{15 x ^{2} ( - 8 x ^{37} y + 1 )}{3 x ^{40} + 2 y ^{3}}

\frac{15 x ^{2} ( - 8 x ^{37} y + 1 )}{3 x ^{40} + 2 y ^{3}}

\int \frac{x ^{3}}{1 - x ^{2}} d x

x \to 4 - lim (2 - x - x^{2})

f (x) = (x^{3} + 9) e^{x}

\int_{e}^{e^{3}} \frac{15}{y} d y

\int x^{2} cos^{2} (x) d x

What is the slope of-10x^3+6x^{40}y+y^4=3 ?
The slope of-10x^3+6x^{40}y+y^4=3 is (15x^2(-8x^{37}y+1))/(3x^{40)+2y^3}