What is the tangent of f(x)=5*3^x,\at x=2 ?

The tangent of f(x)=5*3^x,\at x=2 is y=45ln(3)x+45-90ln(3)

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tangent of f(x)=5*3^x,\at x=2

tangent of

f (x) = 5 \cdot 3^{x}, at x = 2

y = 45 ln (3) x + 45 - 90 ln (3)

Solution steps

Find the tangent point:

(2, 45)

Find the slope of

f (x) = 5 \cdot 3^{x} : \frac{df ( x )}{d x} = 5 \cdot 3^{x} ln (3)

Compute the slope of

f (x) = 5 \cdot 3^{x}

x = 2 : m = 45 ln (3)

Find the line with slope m=

45 ln (3)

and passing through

(2, 45) : y = 45 ln (3) x + 45 - 90 ln (3)

y = 45 ln (3) x + 45 - 90 ln (3)

\frac{d}{d x} (x^{x} + 1)

\frac{\partial}{\partial x} (5 x^{2} + y + sin (a x y) + 3)

\int \frac{3 x ( 8 - x )}{135} d x

y = 2^{x} ln (5)

\int \frac{x}{y} + \frac{y}{x} d y

What is the tangent of f(x)=5*3^x,\at x=2 ?
The tangent of f(x)=5*3^x,\at x=2 is y=45ln(3)x+45-90ln(3)