How do I differentiate this equation for a test tomorrow?

[Morphayne]

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.

 

[sutupidmath]

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

$$y=(x-1)^26^x$$ 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:

$$y'=[(x-1)^26^x]'=[(x-1)^2]'6^x+(6^x)'(x-1)^2=2(x-1)6^x+6^xln6(x-1)^2=(x-1)6^x(2+xln6-ln6)$$

 

[Gib Z]

This question is so much nicer and faster if you use a special trick called Logarithmic differentiation:
If $$f(x) = g(x) \cdot h(x)$$ Then $$\log f(x) = \log(g(x) h(x) ) = \log (g(x)) + \log (h(x))$$ and so $$\frac{ f'(x)}{f(x)} = \frac{g'(x)}{g(x)} + \frac{h'(x)}{h(x)}$$.

So in this case:

$$\log y = \log (x-1)^2 + \log (6^x) = 2\log (x-1) + x \log 6$$

$$\frac{y'}{y} = \frac{2}{x-1} + \log 6$$

Now just multiply through by y.

 

[sutupidmath]

Just in case the op isn't familiar how to differentiate logs in general:

$$log_ax=y<=>a^y=x$$

Lets differentiate implicitly the last part:

$$y'a^ylna=1=>y'=\frac{1}{a^ylna}=\frac{1}{xlna}$$

Now, if a=e then ln(e)=1.