find an equation of the tangent line to the
graph of the function at the given point.
f(x)e^3 1nx,(1,0)

Question

find an equation of the tangent line to the
graph of the function at the given point.
f(x)e^3 1nx,(1,0)

anonymous · Answer

differentiate and sub in x =1 and y = 0

campbell_st · Answer

can you find the 1st derivative

anonymous · Answer

f(x) = e^3x ^ ln(x) .... product rule is needed dy/dx = dv/dx * u + du/dx * v

let u = e^3x and v = ln(x) ==> therefore du/dx = 3e^3x and dv/dx = 1/x

so therefore dy/dx = (e^3x)/x + 3e^3x*lnx

so e^3 + 3e^3*ln1 should equal the gradient

anonymous · Answer

@Iamgmg90 your amazing thanks! (:

anonymous · Answer

of the tangent at that point, which is equal to m

use the formula y_2 - y_1 = m(x_2 - x_1) for x = 1 and y = 0 to calculate the formula, you must be given another coordinate, or be able to infer one

campbell_st · Answer

just a quick question  is the function 

1. $$f(x) = e^{3\ln(x)}$$

or 

2. $$f(x) = e^{3x^{\ln(x)}}$$

as there will be a considerable difference in the derivative.

anonymous · Answer

@campbell_st http://learn.flvs.net/educator/temp/atucker69/2579/ChapterFive/5.04.pdf put that in the search bar its number 56

campbell_st · Answer

its just that 

$$f(x) = e^{3\ln(x)} = e^{\ln(x^3)} = x^3$$

using the log laws.... 

just a thought... 

so the question is 

$$f(x) = e^3 \ln(x) $$

so e^3 is a constant so the derivative is 

$$f'(x) = \frac{e^3}{x}$$

as I said... just a thought...

anonymous · Answer

@campbell_st thank you (: