# Find the value of y' for the function y= ln[(4x^3 - x^2 + 3)e^3x] at x=0.This is a Calculus question

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### 2 Answers

Find the value of `y'` at `x=0` for `y=ln[(4x^3-x^2+3)e^(3x)]` :

First, note that `ln(AB)=lnA+lnB` , so rewrite as:

`y=ln(4x^3-x^2+3)+ln(e^(3x))` or

`y=ln(4x^3-x^2+3)+3x` since `ln,e` are inverses. Then since the derivative of a sum is the sum of derivatives we have:

`y'=1/(4x^3-x^2+3)*(12x^2-2x)+3` or

`y'=(2x(6x-1))/(4x^3-x^2+3)+3`

Evaluating at `x=0` we see that `y'=3` .

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**The value of `y'` when `x=0` for `y=ln[(4x^3-x^2+3)e^(3x)]` is 3.**

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The graph of y and the tangent line y=3x+ln(3):

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**Sources:**

## y= ln[(4x^3 - x^2 + 3)e^3x]

If y=lnf(x) then y'=[1/f(x)]*f'(x)

Here f(x) =(4x^3 - x^2 + 3)e^3x

Now if you consider f(x)= g(x)*p(x) ;

Then f'(x) = g(x)*p'(x)+g'(x)*p(x)

Here g(x) = (4x^3 - x^2 + 3) and p(x)= e^3x

So g'(x) = 4*3*x^2-2*x= 12x^2-2x

Then p'(x) = 3*e^3x

S;

f'(x)=g(x)*p'(x)+g'(x)*p(x)

=(4x^3 - x^2 + 3)*3*e^3x+(12x^2-2x)*e^3x

Finally what we want is the value of y' at x=0

Now

y=ln(fx)

y'= [1/f(x)]*f'(x)

but f(x)=g(x)*p(x) and f'(x)=g(x)*p'(x)+g'(x)*p(x)

So y' = [1/(g(x)*p(x))]*[g(x)*p'(x)+g'(x)*p(x)]

Since x=0;

y' = [1/(g(0)*p(0))]*[g(0)*p'(0)+g'(0)*p(0)]

g(0) = (4*0^3 - 0^2 + 3) = 3

p(0) = e^3*0 = e^0 = 1

g'(0)= 12*0^2-2*0 = 0

p'(0)= 3*e^3*0 = 3e^0 = 3

y' at (x=0) =[1/(g(0)*p(0))]*[g(0)*p'(0)+g'(0)*p(0)]

= [1/(3*1)]*[3*3+0*3]

= (1/3)*9

= 3

**So at x=0; y'=3**