1)
 & =\frac{z^{2}(4+3z^{2}-4z^{3})-(4z+z^{3}-z^{4})(2z)}{(z^{2})^{2}}\:\left(Quotient\: rule\:\frac{dy}{dx}=\frac{vu'-uv'}{v^{2}},\: u=4z+z^{3}-z^{4}\:\&\: v=z^{2}\right)<br />\\ & =\frac{4z^{2}+3z^{4}-4z^{5}-8z^{2}-2z^{4}+2z^{5}}{z^{4}}<br />\\ & =\frac{z^{2}(4+3z^{2}-4z^{3}-8-2z^{2}+2z^{3})}{z^{4}}<br />\\ & =\frac{-2z^{3}+z^{2}-4}{z^{2}}<br />\end{alignedat})
5)
 & =\frac{x^{2}(2)-(2x-4)(2x)}{(x^{2})^{2}}<br />\\ & =\frac{2x^{2}-4x^{2}+8x}{x^{4}}<br />\\ & =\frac{-2x^{2}+8x}{x^{4}}<br />\\ & =\frac{x(-2x+8)}{x^{4}}<br />\\ & =\frac{8-2x}{x^{3}}<br />\end{alignedat})
So next you would have to find where the curve crosses the x-axis, i.e.
, so solve
for x.
Now you have the point
)
Find the gradient by subbing that into your derivative (you should be able to do the working for that part).
6. For this, you have two unknowns, so you need 2 equations. One will come from subbing in the point
)
(you should get
). The other will come from finding the derivative (it should be
, then letting that equal the gradient (7), and substituting in the x-value (3). You should get
.
For the last one, I'm assuiming its
?
Firstly find the derivative (so that you have something with the gradient you can play with). (Remember that you can also write
like
, this may help you when you take the derivative, getting the powers right). So we have
.
Now you can substitute the
into the derivative to find the gradient at that point (in terms of k), do the same for
. Now you have two gradients in terms of k, and we know that they are perpendicular, so that means they satisfy
, so sub them into the equ and solve for k.
If you need anything more explained then post back, hope it helps
EDIT: For some reason I said it was the chain rule instead of the Quotient rule, fixed it.
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