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

The tangent of f(x)=ln(e^{x^2}),\at x=1 is y=2x-1

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tangent of f(x)=ln(e^{x^2}),\at x=1

tangent of

f (x) = ln (e^{x^{2}}), at x = 1

y = 2 x - 1

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

f (x) = ln (e^{x^{2}})

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 1) : y = 2 x - 1

y = 2 x - 1

\int \frac{2 x + 5}{x ^{2} + 2 x + 5} d x

\int_{1}^{4} (x^{2} - 4 x - 3) d x

\int cos (6 θ) d θ

x \to \frac{1}{5} lim (\frac{x}{( 10 x - 2 )})

\int \frac{1}{( x ^{2} + 3 ) ^{2}} d x

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