What is the tangent of y=(1+5x)^9,(0,1) ?

The tangent of y=(1+5x)^9,(0,1) is y=45x+1

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tangent of y=(1+5x)^9,(0,1)

tangent of

y = (1 + 5 x)^{9}, (0, 1)

y = 45 x + 1

Solution steps

Find the slope of

y = (1 + 5 x)^{9} : \frac{d y}{d x} = 45 (1 + 5 x)^{8}

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 = 45

Find the line with slope m=

45

and passing through

(0, 1) : y = 45 x + 1

y = 45 x + 1

y^{2} y^{^{'}} + 5 = x^{2}

\int \frac{7 x - 5}{x ^{2} + 16} d x

f (x) = 6 - x^{2}, at x = 7

\int \frac{1}{tan ( x ) - 1} d x

\int_{- 5}^{6} x d x

What is the tangent of y=(1+5x)^9,(0,1) ?
The tangent of y=(1+5x)^9,(0,1) is y=45x+1