c) is the derivative of a)f) is the derivative of d)b) is the derivitive of h)

andg)ande)are irrelevant

Assuming u know:When the derivative function = 0 at a pointx, thenyatxis a turning point (trough or a crest)

first one because the turning points ofa),(ie when gradient is 0), at at thex-intersections ofc). This is the same for second one.

How ever you might think thate)is the derivative ofc):

-the thing is thatyvaluee)goes fromnegative to positivein the first intersection (from left)

-and the first (from left) turning point is a crest, which goes frompositive and negativegradient

- soe)is not the derivative ofc)

If you integratec)you get a power-4 graph, and we've used the only one

If you differentiate it, you get a straight line: which there is none

So we are left withb), g)andh):

Personally, Ive never seen a graph like h) before.

But this is what i think:

lets divide the graph curves into positiveand negativex(right)x(left)

considerb)having a derivative

left: the gradient is positive

right: the gradient is positive

both other graphs have curves negative y, sob)doesnot have a derivative.

Considerg)

left:gradient is negative

right: gradient is positive

yvalues ofh)are always negative andb)is positive on theleftandnegativeon the right (opposite)

Sog)also does not have a derivative

Soh)does have one

gradient

-left: positive

-right:negative

-also whenxapproaches0, the gradient is infinitely negative/positive

And these properties are shown in the curveb),and so... the third pair.

Now Ive said a bit tooo much