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

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

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tangent of y=(x+1)^{x^2},(0,1)

tangent of

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

y = 1

Solution steps

Find the slope of

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

Compute the slope of

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

(0, 1) : m = 0

Find the line with slope m=

0

and passing through

(0, 1) : y = 1

y = 1

\int \frac{3}{3 x + 2} d x

\int_{1}^{2} (x^{3}) d x

\frac{d}{d x} (7 - (1.01)^{- x^{2}})

f (x) = x^{2} + 1, (- 5, 26)

\frac{\partial}{\partial x} (3 x^{2} - 6 x - 9)

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