What is the tangent of f(x)=3^{x^2-1},\at x=2 ?

The tangent of f(x)=3^{x^2-1},\at x=2 is y=108ln(3)x+27-216ln(3)

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tangent of f(x)=3^{x^2-1},\at x=2

tangent of

f (x) = 3^{x^{2} - 1}, at x = 2

y = 108 ln (3) x + 27 - 216 ln (3)

Solution steps

Find the tangent point:

(2, 27)

Find the slope of

f (x) = 3^{x^{2} - 1} : \frac{df ( x )}{d x} = 2 ln (3) \cdot 3^{x^{2} - 1} x

Compute the slope of

f (x) = 3^{x^{2} - 1}

x = 2 : m = 108 ln (3)

Find the line with slope m=

108 ln (3)

and passing through

(2, 27) : y = 108 ln (3) x + 27 - 216 ln (3)

y = 108 ln (3) x + 27 - 216 ln (3)

y^{^{''}} - y = 4 sin (3 t)

\int - 2 x^{4} - 4 x^{2} - 5 d x

\frac{d}{d x} (cot (\frac{π}{2}))

\frac{d ^{2} y}{d x ^{2}} (x^{2}) - 9 x \frac{d y}{d x} + (61 y) = 0

x \to - 1 lim (2 x^{2} + 3 x - 5)

What is the tangent of f(x)=3^{x^2-1},\at x=2 ?
The tangent of f(x)=3^{x^2-1},\at x=2 is y=108ln(3)x+27-216ln(3)