As you can see, the gradient is always twice the value of its original X value Where y=x². So the gradient function has to be f `(x)=2x for y=x².
I will now move on to the y=2x² graph and find its gradient function. I will draw the graph (page 4) and put my findings in a table. I will now find the gradients for a selected number of points in the graph y=2x² so as to find the gradient function:
From the table, you can see that the gradient of the point is always 4 times that of the amount of X. Which means that the gradient function for y=2x² is f ‘(x)=4x.
Now that I have found both the gradient functions for y=x² and y=2x², I shall draw the y=ax² graph (page 5) to show the equation used to find the gradient function for y=x² graphs. I will use a table to present my findings:
As you can see, this information above is very similar to y=x² ‘s information. Its is simply adding an ‘a’ to the Y-axis. But nevertheless, you can see that the gradient function for y=ax² is f ‘(x)=2ax, where ‘a’ is a constant. This gradient function can be used for any graph as long as it is a similar graph to y=x². i.e. y=4x², y=10x² or even y=250x².
The graph y=x² is a rather simple one, when trying to find the gradient function. That is why I will use the graph y=x³ and y=2x³ to find the gradient function of y=ax³, and hopefully be able to find the gradient function for all graphs, namely y=axⁿ. I will, obviously start with y=x³ and shall use a program called, “autograph”, to draw the graphs for me from now on as the graphs are noticeably more accurate than drawing by hand.
The gradient is very inconsistent as it increases or decreases and it is probably due to the fact that the gradient function has a power in it. After using trial and improvement method, I found out that the gradient function for y=x³ is f ‘(x)=3x².
The next graph I will be using is y=2x³. the graph will be on the next page as it cannot fit on this current one, but I will put the table below.
Since this graph is just twice the amount of y=x³, I knew that it had to be twice of f ‘(x)=3x², which is f ‘(x)=6x². I was right, and the gradient function for y=2x³ was easier to find than y=x³ as I already knew the gradient function for y=x³, which decreased the time spent on finding its gradient function. Now, all we have to do is the graph of y=ax³ and then generalise y=axⁿ. The graph y=ax³ is pretty close to y=x³, as y=ax² is to y=x². As always, I will place the graph and the table of information below and find the gradient function.
Using the graph and the table in the previous page, I can tell that the gradient function for y=ax³ is f ‘(x)=3ax².
Now that I have done two different types of graphs, I have noticed some things. The gradient function is always the constant (a), times by ‘x’, and then added with a power after ‘x’. The power is also always one number lower than the equation of the graph.
Using my observations, I will now predict what the gradient function for y=ax4. I predict that the gradient function of y=ax4 will be f ‘(x)=4ax³.
I then used my graph to find the gradient, here are my results:
This proves that my prediction is right. This means that if I generalise y=axⁿ, the gradient function would be f ‘(x)=anx(n-1).
I can also use differentiation to find the gradient function for any graph and also to generalise y=axⁿ. The next page shows how I go about doing it.
Using graphs to find the gradient function is not perfect and therefore the calculations may be wrong. I will use differentiation to work out the gradient on a consistent basis. Observe the picture below:
You should be able to see that the gradient of the chord (in red) will be the gradient of the function, as h tends to zero.
To find the gradient function of any curve, the equation is:
f(x+h) - f(x)
(x+h) - x
= ( f(x+h) - f(x) )
h
Which means that the gradient of the function at x should be the limit, as h tends to zero.
When we substitute x² into the equation, the equation would be:
f ‘(x)= (x+h)²-x²
h
= x²+2xh+h²-x²
h
= 2xh+h²
h
f ‘(x)= 2x+h
But if h tends to 0, h=0, then it would be
f ‘(x)= 2x +0 = 2X, where y=x²
If I substituted y=x³ into the equation, I would get f ‘(x)=3x². If I did the same for y=x4, it would be f ‘(x)=4x³. This now shows a very obvious link which would help me generalise y=axn. In the equation, the square would move down in front of ‘x’ and the square of ‘x’ would be a digit lower than the original square. This means that the gradient function of y=axn is f ‘(x)=anx(n-1) .