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

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

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

tangent of

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

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

Compute the slope of

y = 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)

\int 3^{5 x - 2} d x

\frac{d}{d x} (\frac{800}{x})

x \to 4 lim (\frac{2}{x - 4} - \frac{x + 4}{x ^{2} - 4 x})

\int \frac{x ^{5} - x ^{3} + 4 x}{x ^{4}} d x

\int \frac{sin ( y )}{y} d y

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