What is the tangent of y=(3x+1/x)^2,(1,16) ?

The tangent of y=(3x+1/x)^2,(1,16) is y=16x

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tangent of y=(3x+1/x)^2,(1,16)

tangent of

y = (3 x + \frac{1}{x})^{2}, (1, 16)

y = 16 x

Solution steps

Find the slope of

y = (3 x + \frac{1}{x})^{2} : \frac{d y}{d x} = 18 x - \frac{2}{x ^{3}}

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 = 16

Find the line with slope m=

16

and passing through

(1, 16) : y = 16 x

y = 16 x

\int sec^{2} (x) (cos^{3} (x) + 1) d x

x \to 1 + lim (ln (x - 1))

\int_{0}^{\infty} e^{- 4 t} d t

π^{4}

x \to \infty lim (x - 2)

What is the tangent of y=(3x+1/x)^2,(1,16) ?
The tangent of y=(3x+1/x)^2,(1,16) is y=16x