What is the tangent of y=(1+5x)^{10},(0,1) ?

The tangent of y=(1+5x)^{10},(0,1) is y=50x+1

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tangent of y=(1+5x)^{10},(0,1)

tangent of

y = (1 + 5 x)^{10}, (0, 1)

y = 50 x + 1

Solution steps

Find the slope of

y = (1 + 5 x)^{10} : \frac{d y}{d x} = 50 (1 + 5 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 = 50

Find the line with slope m=

50

and passing through

(0, 1) : y = 50 x + 1

y = 50 x + 1

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

\int (x + 2)^{11} d x

y^{^{''}} + 2 y^{^{'}} + 5 y = 9 sin (2 t)

\int ln ∣ x + 1 ∣ d x

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

What is the tangent of y=(1+5x)^{10},(0,1) ?
The tangent of y=(1+5x)^{10},(0,1) is y=50x+1