What is the tangent of y=x^3+7x-5,(1,3) ?

The tangent of y=x^3+7x-5,(1,3) is y=10x-7

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tangent of y=x^3+7x-5,(1,3)

tangent of

y = x^{3} + 7 x - 5, (1, 3)

y = 10 x - 7

Solution steps

Find the slope of

y = x^{3} + 7 x - 5 : \frac{d y}{d x} = 3 x^{2} + 7

Compute the slope of

y = x^{3} + 7 x - 5

(1, 3) : m = 10

Find the line with slope m=

10

and passing through

(1, 3) : y = 10 x - 7

y = 10 x - 7

\int \frac{8 x ^{3}}{( x ^{2} + 2 ) ^{2}} d x

\int (\frac{8}{x}) d x

\frac{d y}{d x} = \frac{1}{x ^{8}}

y = (3 - 2 x + 4 x^{3}), (- 1, 1)

\frac{d}{d x} (x ln (y) + sin (3 x))

What is the tangent of y=x^3+7x-5,(1,3) ?
The tangent of y=x^3+7x-5,(1,3) is y=10x-7