What is the normal of f(x)=x*5^x,\at x=1 ?

The normal of f(x)=x*5^x,\at x=1 is f(x)=-1/(5+5ln(5))x+5+1/(5+5ln(5))

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normal of f(x)=x*5^x,\at x=1

normal of

f (x) = x \cdot 5^{x}, at x = 1

f (x) = - \frac{1}{5 + 5 ln ( 5 )} x + 5 + \frac{1}{5 + 5 ln ( 5 )}

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 5 + 5 ln (5)

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{5 + 5 ln ( 5 )}

Find the line with slope m=

- \frac{1}{5 + 5 ln ( 5 )}

and passing through

(1, 5) : f (x) = - \frac{1}{5 + 5 ln ( 5 )} x + 5 + \frac{1}{5 + 5 ln ( 5 )}

f (x) = - \frac{1}{5 + 5 ln ( 5 )} x + 5 + \frac{1}{5 + 5 ln ( 5 )}

What is the normal of f(x)=x*5^x,\at x=1 ?
The normal of f(x)=x*5^x,\at x=1 is f(x)=-1/(5+5ln(5))x+5+1/(5+5ln(5))