What is the tangent of t^3+1t,\at t=1 ?

The tangent of t^3+1t,\at t=1 is y=4t-2

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tangent of t^3+1t,\at t=1

tangent of

t^{3} + 1 t, at t = 1

y = 4 t - 2

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

y = t^{3} + 1 t : \frac{d y}{d t} = 3 t^{2} + 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 = 4

Find the line with slope m=

4

and passing through

(1, 2) : y = 4 t - 2

y = 4 t - 2

\frac{\partial}{\partial u} (3 y uxu v^{3} - 2)

\int \frac{1}{100 - 2 x} d x

\frac{\partial}{\partial x} ((3 x^{2} + 4)^{3})

\frac{d y}{d x}, sec (9 x + 2) + 6 y^{2} = 42

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