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

\frac{6}{3 x ^{2}} - x, at x = 1

y = - 5 x + 10

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

y = \frac{6}{3 x ^{2}} - x : \frac{d y}{d x} = - \frac{4 x}{( x ^{2} ) ^{\frac{4}{3}}} - 1

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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

d er i v a t i v e f (x) = (2 - 3 x)^{3}

\int g \cdot \frac{m}{k} (1 - e^{- \frac{k}{m} \cdot t}) d t

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