What is the tangent of 4x^2=ln(x-y)+3,\at x=0 ?

The tangent of 4x^2=ln(x-y)+3,\at x=0 is y=x-1/(e^3)

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tangent of 4x^2=ln(x-y)+3,\at x=0

tangent of

4 x^{2} = ln (x - y) + 3, at x = 0

y = x - \frac{1}{e ^{3}}

Solution steps

Find the tangent point:

(0, - \frac{1}{e ^{3}})

Find the slope of

4 x^{2} = ln (x - y) + 3 : \frac{d y}{d x} = - e^{4 x^{2} - 3} \cdot 8 x + 1

Compute the slope of

4 x^{2} = ln (x - y) + 3

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, - \frac{1}{e ^{3}}) : y = x - \frac{1}{e ^{3}}

y = x - \frac{1}{e ^{3}}

\int e^{x} (4 + e^{x})^{5} d x

(x^{''} + x)^{^{'}}

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

\int_{6}^{8} \frac{94}{( x - 6 ) ^{3}} d x

(t + 5)^{\frac{2}{3}} (2 t^{2} - 3)^{3}

What is the tangent of 4x^2=ln(x-y)+3,\at x=0 ?
The tangent of 4x^2=ln(x-y)+3,\at x=0 is y=x-1/(e^3)