0
$\begingroup$

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.

1 Answers 1

1

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:

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

  • 0
    Wow I $f$eel like an idiot, I multiplied out the brackets!. Thank you very much.2012-06-02