What is the tangent of x^6+y^2=2,(1,-1) ?

The tangent of x^6+y^2=2,(1,-1) is y=3x-4

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tangent of x^6+y^2=2,(1,-1)

tangent of

x^{6} + y^{2} = 2, (1, - 1)

y = 3 x - 4

Solution steps

Find the slope of

x^{6} + y^{2} = 2 : \frac{d y}{d x} = - \frac{3 x ^{5}}{y}

Compute the slope of

x^{6} + y^{2} = 2

(1, - 1) : m = 3

Find the line with slope m=

3

and passing through

(1, - 1) : y = 3 x - 4

y = 3 x - 4

\frac{d}{d x} (2 A + B (e^{x} x^{2} + 4 e^{x} x + 2 e^{x}))

x y^{^{'}} + 2 y = 3 cos (x)

\int cos^{2} (2 x) sin^{3} (2 x) d x

h (x) = 4^{x} - 6

g (x) = 3 x^{2} cos (x) + 8 x

What is the tangent of x^6+y^2=2,(1,-1) ?
The tangent of x^6+y^2=2,(1,-1) is y=3x-4