What is the tangent of f(x)=ln(1+2x^5),(0,0) ?

The tangent of f(x)=ln(1+2x^5),(0,0) is y=0

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tangent of f(x)=ln(1+2x^5),(0,0)

tangent of

f (x) = ln (1 + 2 x^{5}), (0, 0)

y = 0

Solution steps

Find the slope of

f (x) = ln (1 + 2 x^{5}) : \frac{df ( x )}{d x} = \frac{10 x ^{4}}{1 + 2 x ^{5}}

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 = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

f (x) = \frac{1 + 3 x}{x - 2}

\frac{d}{d x} (lo g_{b} (x))

\int e^{d} x d x

f (x) = \frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}

x \to 3 lim (x - 5)

What is the tangent of f(x)=ln(1+2x^5),(0,0) ?
The tangent of f(x)=ln(1+2x^5),(0,0) is y=0