What is the tangent of y=7x-x^2,\at x=1 ?

The tangent of y=7x-x^2,\at x=1 is y=5x+1

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tangent of y=7x-x^2,\at x=1

tangent of

y = 7 x - x^{2}, at x = 1

y = 5 x + 1

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

y = 7 x - x^{2} : \frac{d y}{d x} = 7 - 2 x

Compute the slope of

y = 7 x - x^{2}

x = 1 : m = 5

Find the line with slope m=

5

and passing through

(1, 6) : y = 5 x + 1

y = 5 x + 1

f (x) = 3^{c o s (x)}

\int tan (x) d x

\frac{d}{d x} (\frac{1 - e ^{x}}{1 + e ^{x}})

(ln (x^{'}))^{^{'}}

f (x) = 4 sin (2 x) + \frac{2 x ^{5} - x}{x}

What is the tangent of y=7x-x^2,\at x=1 ?
The tangent of y=7x-x^2,\at x=1 is y=5x+1