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Given that: f(−4)=4, f′(−4)=3, g(−4)=8, g′(−4)=9, calculate the following: (fg)′(−4)= (f/g)′(−4)= (g/f)′(−4)=
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@mathmale
By product rule we have,\[\large\rm (fg)'=f'g+fg'\]Evaluated at x=-4 gives us,\[\large\rm (fg)'(-4)=f'(-4)g(-4)+f(-4)g'(-4)\]Ya? :) Any confusion there?
a little
oh wait
(3*8)+(4*9)
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is that right?
Good good good.
For the others, remember your quotient rule?
a little bit
Medals 1 ((3*8)-(4*9))/(8^2)
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?
\[\large\rm \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}\] Mmm yes looks good!
how would you do the last one thought, im a little confused?
Umm you just follow your rule for quotient.\[\large\rm \left(\frac{g}{f}\right)'=\frac{g'f-gf'}{f^2}\]Dtop*bottom minus Dbottom*top all divided by bottom^2
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