What is the tangent of f(x)=x(36+x^2)^{-1/2},\at x=0 ?

The tangent of f(x)=x(36+x^2)^{-1/2},\at x=0 is y= 1/6 x

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tangent of f(x)=x(36+x^2)^{-1/2},\at x=0

tangent of

f (x) = x (36 + x^{2})^{- \frac{1}{2}}, at x = 0

y = \frac{1}{6} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = x (36 + x^{2})^{- \frac{1}{2}} : \frac{df ( x )}{d x} = \frac{36}{( x ^{2} + 36 ) ^{\frac{3}{2}}}

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{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(0, 0) : y = \frac{1}{6} x

y = \frac{1}{6} x

113 t - 0.5 t^{4} + 900

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

\frac{\partial}{\partial s} (3 s + t)

f (x) = 3 csc (x) - 5 sec (x)

\int \frac{x + 2}{x ^{2} + 4 x + 1} d x

What is the tangent of f(x)=x(36+x^2)^{-1/2},\at x=0 ?
The tangent of f(x)=x(36+x^2)^{-1/2},\at x=0 is y= 1/6 x