# Help with differentiation.

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)

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
Homework Helper
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.

Last edited:
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.