What is the tangent of ln(x^2-24)=x-y+6,(-5,1) ?

The tangent of ln(x^2-24)=x-y+6,(-5,1) is y=11x+56

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tangent of ln(x^2-24)=x-y+6,(-5,1)

tangent of

ln (x^{2} - 24) = x - y + 6, (- 5, 1)

y = 11 x + 56

Solution steps

Find the slope of

ln (x^{2} - 24) = x - y + 6 : \frac{d y}{d x} = - \frac{2 x}{x ^{2} - 24} + 1

Compute the slope of

ln (x^{2} - 24) = x - y + 6

(- 5, 1) : m = 11

Find the line with slope m=

11

and passing through

(- 5, 1) : y = 11 x + 56

y = 11 x + 56

\int (x^{2} - 5 x + 5) d x

y^{^{''}} + y = - sin (2 t)

\frac{d}{d x} (\frac{5}{4 x - 3})

\int 2 (sin (x) - cos (x) + 1) d x

tan (5 x)

What is the tangent of ln(x^2-24)=x-y+6,(-5,1) ?
The tangent of ln(x^2-24)=x-y+6,(-5,1) is y=11x+56