What is the tangent of f(x)=sqrt(18x+18),(1,6) ?

The tangent of f(x)=sqrt(18x+18),(1,6) is y= 3/2 x+9/2

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tangent of f(x)=sqrt(18x+18),(1,6)

tangent of

f (x) = 18 x + 18, (1, 6)

y = \frac{3}{2} x + \frac{9}{2}

Solution steps

Find the slope of

f (x) = 18 x + 18 : \frac{df ( x )}{d x} = \frac{9}{18 x + 18}

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 = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(1, 6) : y = \frac{3}{2} x + \frac{9}{2}

y = \frac{3}{2} x + \frac{9}{2}

(2 x y + 2 y^{2}) d x + (x^{2} + 4 x y) d y = 0

d er i v a t i v e x^{3} (x - 2)^{2}

\int 27 x^{2} d x

\int 6 4 x d x

\int_{- \frac{π}{4}}^{0} 4 sec^{3} (x) d x

What is the tangent of f(x)=sqrt(18x+18),(1,6) ?
The tangent of f(x)=sqrt(18x+18),(1,6) is y= 3/2 x+9/2