What is the tangent of 6/(\sqrt[3]{x^2)}-x,\at x=1 ?

The tangent of 6/(\sqrt[3]{x^2)}-x,\at x=1 is y=-5x+10

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tangent of 6/(\sqrt[3]{x^2)}-x,\at x=1

tangent of

y = - 5 x + 10

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

Compute the slope of

x = 1 : m = - 5

Find the line with slope m=

- 5

and passing through

(1, 5) : y = - 5 x + 10

y = - 5 x + 10

\frac{d y}{d x} = 17

\frac{\partial}{\partial x} (y cos (π x))

\int t^{6} ln (t) d t

f (x) = (2 - 3 x)^{3}

\int sin (x) - 4 cos (x) + 1 d x

What is the tangent of 6/(\sqrt[3]{x^2)}-x,\at x=1 ?
The tangent of 6/(\sqrt[3]{x^2)}-x,\at x=1 is y=-5x+10