What is the tangent of y=(1-3x)^{10},(0,1) ?

The tangent of y=(1-3x)^{10},(0,1) is y=-30x+1

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tangent of y=(1-3x)^{10},(0,1)

tangent of

y = (1 - 3 x)^{10}, (0, 1)

y = - 30 x + 1

Solution steps

Find the slope of

y = (1 - 3 x)^{10} : \frac{d y}{d x} = - 30 (1 - 3 x)^{9}

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 = - 30

Find the line with slope m=

- 30

and passing through

(0, 1) : y = - 30 x + 1

y = - 30 x + 1

\int e^{5 - x} d x

\int_{0}^{1} (2 x - 1)^{6} d x

\int \frac{1}{( 5 x - 2 ) ^{\frac{3}{2}}} d x

x \to 2 lim (sin (x - 2) + 1)

\frac{d}{d x} ((x - 1)^{2} (x - 2)^{3})

What is the tangent of y=(1-3x)^{10},(0,1) ?
The tangent of y=(1-3x)^{10},(0,1) is y=-30x+1