# Implicit Differentiation Step Explanation

I'm trying to follow some notes on beam mechanics, and there is a differentiation step I don't understand.

He goes from:

$M_n(\lambda(1-R)+R) = \frac{1}{3} + \frac{(R-1)(1-\lambda)^2(2+\lambda)}{6}$

to:

$\frac{dM_n}{d\lambda}(\lambda(1-R)+R) + M_n(1-R) = \frac{R-1}{6}(-2(1-\lambda)(2+\lambda)+(1-\lambda)^2)$

I can do the R.H.S, but I don't understand how the L.H.S is differentiated. Also some insights into how the R.H.S is differentiated in one step would be very useful!

Can anyone help me understand this please?

Thanks.

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Consider the lhs as a product of two functions $f(\lambda) = M_n (\lambda)$ and $g(\lambda) = \lambda(1-R) +R$.

Note that $f'(\lambda) = \frac{dM_n}{d\lambda}$, $g'(\lambda) = 1-R$.

Then the derivative wrt $\lambda$ is:

$$f'(\lambda)g(\lambda)+f(\lambda)g'(\lambda)= \frac{dM_n}{d\lambda}(\lambda(1-R) +R)+M_n (\lambda)(1-R).$$

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Ah of course, thank you. It's been a long time since I've done any of this. You've got a slight typo though the (1+R) terms should be (1-R) –  Griffin Jun 2 '12 at 16:22
Any idea how the R.H.S is done in one step, btw? –  Griffin Jun 2 '12 at 16:24
I am not sure what you mean by one step. It is another application of the product rule applied to $(1-\lambda)^2(2+\lambda)$. –  copper.hat Jun 2 '12 at 16:41
Wow I feel like an idiot, I multiplied out the brackets!. Thank you very much. –  Griffin Jun 2 '12 at 16:55