Help with differentiation.

  • Thread starter Morphayne
  • Start date
  • #1
Morphayne
13
0
I need help differentiating this equation:

y=(x-1)^2(6^x)

What I have (I went wrong somewhere):

dy/dx = (x-1)[(x-1)(6^x)]

=(x-1)[6^x(1+ x ln6 - ln6)

=(x-1)6^x(1+ x ln6 - ln6)

So;

My final answer: (x-1)6^x(1+ x ln6 - ln6)

Answer in the textbook: (x-1)6^x(2+ x ln6 - ln6)

Please help. I need to understand this for a test tomorrow.
 

Answers and Replies

  • #2
sutupidmath
1,631
4
y=(x-1)^2(6^x)

[tex]y=(x-1)^26^x[/tex] I am assuming this is what you want to differentiate with respect to x, right?

First the big structure tells us that we first need to use product rule, after that chain rule and so on, so:

[tex]y'=[(x-1)^26^x]'=[(x-1)^2]'6^x+(6^x)'(x-1)^2=2(x-1)6^x+6^xln6(x-1)^2=(x-1)6^x(2+xln6-ln6)[/tex]
 
  • #3
Gib Z
Homework Helper
3,352
6
This question is so much nicer and faster if you use a special trick called Logarithmic differentiation:
If [tex]f(x) = g(x) \cdot h(x)[/tex] Then [tex]\log f(x) = \log(g(x) h(x) ) = \log (g(x)) + \log (h(x)) [/tex] and so [tex] \frac{ f'(x)}{f(x)} = \frac{g'(x)}{g(x)} + \frac{h'(x)}{h(x)}[/tex].

So in this case:

[tex] \log y = \log (x-1)^2 + \log (6^x) = 2\log (x-1) + x \log 6[/tex]

[tex] \frac{y'}{y} = \frac{2}{x-1} + \log 6[/tex]

Now just multiply through by y.
 
Last edited:
  • #4
sutupidmath
1,631
4
Just in case the op isn't familiar how to differentiate logs in general:

[tex] log_ax=y<=>a^y=x[/tex]

Lets differentiate implicitly the last part:

[tex] y'a^ylna=1=>y'=\frac{1}{a^ylna}=\frac{1}{xlna}[/tex]

Now, if a=e then ln(e)=1.
 

Suggested for: Help with differentiation.

  • Last Post
Replies
2
Views
277
Replies
0
Views
96
Replies
10
Views
444
  • Last Post
Replies
5
Views
657
Replies
24
Views
454
Replies
15
Views
394
Replies
3
Views
284
Replies
5
Views
501
  • Last Post
Replies
7
Views
418
Top