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What do these brackets mean Calculus!!

Do the brackets around the equation in part 3(a) (iii) mean that you have to differentiate what is in the brackets first and then apply the chain rule, would you also have to apply the chain rule to the term with the 4ln (7x) in it as well

The equation is included below, thanks

Attachment 23009http://www.mathhelpforum.com/math-he...isc/pencil.png

Re: What do these brackets mean Calculus!!

The brackets are just there to show that the whole expression within is raised to the power $\displaystyle 4$. Are you aware of the general rule for differentiating something of the form $\displaystyle y=[f(x)]^n$?

Re: What do these brackets mean Calculus!!

yeah so its goin to be a chain rule within a chain rule or is it just going to be a straightforward chain rule equation after you assign what is in the brackets to lets say t and then differentiate that

Re: What do these brackets mean Calculus!!

Oh, you mean you want to solve it via a substitution. You don't have to solve it that way, which is why I was confused. Letting $\displaystyle t=\frac{6}{x}+4\ln(7x)$

And then using the fact that $\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}$ is the way to go. Why do you think you need to use the chain rule twice?

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Re: What do these brackets mean Calculus!!

Would this be the worked answer or am I missing something thanks

Re: What do these brackets mean Calculus!!

Sorry I know the x after the 7 on the last line is missing and the power of 3 is also missing

Re: What do these brackets mean Calculus!!

You start off correctly, but towards the end you get into a ball of confusion.

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$

$\displaystyle y=8t^4$ and $\displaystyle t=6x^{-1}+4\ln(7x)$

$\displaystyle \frac{dy}{dt}=32t^3$

$\displaystyle \frac{dt}{dx}=-6x^{-2}+\frac{4}{x}$

Have another attempt.