What is the tangent of f(x)=arcsin(x)+x^2,\at x=0 ?

The tangent of f(x)=arcsin(x)+x^2,\at x=0 is y=x

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tangent of f(x)=arcsin(x)+x^2,\at x=0

tangent of

f (x) = arcsin (x) + x^{2}, at x = 0

y = x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

f (x) = arcsin (x) + x^{2}

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

\int cos^{4} (7 x + 9) d x

\frac{d}{d x} (2 x^{2} e^{2 x})

\frac{x ^{2}}{1 + 8 x}

\frac{d}{d x} (2 x + 1)

\int_{- \infty}^{0} e^{8 x} d x

What is the tangent of f(x)=arcsin(x)+x^2,\at x=0 ?
The tangent of f(x)=arcsin(x)+x^2,\at x=0 is y=x