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

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

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

tangent of

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

y = 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

f (x) = (x^{2} + 1)^{x + 1}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 1) : y = 1

y = 1

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\int (7 x^{3} - 7)^{4} x^{2} d x

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