What is the tangent of y=x^3-5x+1,(3,13) ?

The tangent of y=x^3-5x+1,(3,13) is y=22x-53

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tangent of y=x^3-5x+1,(3,13)

tangent of

y = x^{3} - 5 x + 1, (3, 13)

y = 22 x - 53

Solution steps

Find the slope of

y = x^{3} - 5 x + 1 : \frac{d y}{d x} = 3 x^{2} - 5

Compute the slope of

y = x^{3} - 5 x + 1

(3, 13) : m = 22

Find the line with slope m=

22

and passing through

(3, 13) : y = 22 x - 53

y = 22 x - 53

f^{^{''}} (x) = \frac{4}{x ^{2}}

\int \frac{1}{x ^{2} 49 x ^{2} + 16} d x

\frac{d}{d t} (e^{t a n (π t)})

\frac{x ^{2}}{e ^{x}}

\int arcsin (x) 1 - x d x

What is the tangent of y=x^3-5x+1,(3,13) ?
The tangent of y=x^3-5x+1,(3,13) is y=22x-53