Please help with differentiation and simplification.

[val1]

Differentiate the following two expressions with respect to x

a)
\(\displaystyle \L

y = \frac{{2 - x}}{{\sqrt {x^2 - 1} }}\cr

{\rm [I used the quotient rule] } \cr

{\rm let }u = 2 - x \cr
\frac{{du}}{{dx}} = - 1 \cr
\cr
{\rm let }v = \left( {x^2 - 1} \right)^{ - {\textstyle{1 \over 2}}} \cr
\frac{{dv}}{{dx}} = - \frac{1}{2}\left( {x^2 - 1} \right)^{ - {\textstyle{3 \over 2}}} \left( {2x} \right) \cr
\cr
\frac{{dy}}{{dx}} = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{v^2 }} \cr
= \frac{{ - \left( {x^2 - 1} \right)^{ - {\textstyle{1 \over 2}}} - \left( {2 - x} \right)\left( { - {\textstyle{1 \over 2}}} \right)\left( {x^2 - 1} \right)^{ - {\textstyle{3 \over 2}}} \left( {2x} \right)}}{{\left( {x^2 - 1} \right)}} \cr
= \frac{{\frac{{ - \left( x \right)\left( {2 - x} \right)}}{{ - \left( {x^2 - 1} \right)^{{\textstyle{1 \over 2}}} \left( {x^2 - 1} \right)^{{\textstyle{3 \over 2}}} }}}}{{\left( {x^2 - 1} \right)}} \cr
\cr
{\rm Is this ok so far and how would I simplify this? } \cr \cr\)




b)
\(\displaystyle \L

y = \left( {2x\sqrt {x^2 + 4x} + \frac{1}{{\sqrt {x^2 + 4x} }}} \right)^2 \\
\\
\\

{\rm How would I differentiate this? Thank you }\)

 

[stapel]

a) To use the Quotient Rule, you have to let:


. . . . .\(\displaystyle \large{y(x)\,=\,\frac{u(x)}{v(x)}}\)


Then:


. . . . .\(\displaystyle \large{v(x)\,=\,\sqrt{x^2\,-\,1}\,=\,\left(x^2\,-\,1\right)^{1/2}}\)


Note the sign on the final exponent. (You would use "to the minus one-half" if you wanted to use the Product Rule.)

Try it using the above for v(x).

b) For this, use the Chain Rule. First, differentiate the "square", which is the "outside" function. Then draw a nice big left-hand open-bracket, and differentiate everything that was inside the square. This will involve the Product Rule, a "plus", and the Quotient Rule. End with a nice big right-hand close-bracket to finish.

I don't know how picky your book is about "simplifying". It's gonna be pretty messy.... :shock:

Eliz.