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The chain rule is used when you're trying to differentiate a complicated function, and it allows you to split up the function into easier ones. The trick is spotting where to split the function, and what part you should substitute.  Say, for example, you're trying to find the derivative of:   y=(1+x)^2  . If you can replace a suitable function of x with a function of u, you would be left with y=f(u) and u=g(x), y and u being functions of u and x respectively. Now, if you differentiate y with respect to u, and u with repsect to x, you're left with the equation:  dy/dx = dy/du * du/dx  . On the RHS, you're left with dy/dx, as the du's cancel out, which is what you were after at the beginning. All that you would have to do is replace any remaining u with your function of x to find your desired derivative.  It's better understood through an example. Let's have a look at the example above, y=(1+x)^2. Take u=1+x, and youre left with y=u^2. Now, differentiate both fuctions, y=f(u) and u=g(x), and you're left with:  dy/du=2u  and  du/dx=1. Now, from the formula above, we know that dy/dx=dy/du * du/dx. Therefore:  dy/dx = 2u * 1 = 2u. Now all that's left to do is replace u with the function of x you originally took out. In this case, u=1+x, so:  dy/dx = 2(1+x) .    For a slightly harder example, try: y=1/(x^2+4x)^1/2 . Rearrange this to have y in terms of powers, similar to the first example: y=(x^2+4x)^-1/2 .  Now, take an appropriate function of x to equal u. In this example, it would be a good choice to take u=x^2+4x , leaving y=u^-1/2. Differentiating both of these functions gives:  dy/du=-1/2(u)^-3/2  and  du/dx=2x+4. We know that dy/dx=dy/du * du/dx, therefore:  dy/dx = -1/2(u)^-3/2 * (2x+4) . Replacing u with the function of x you took out gives:   dy/dx = -1/2(x^2+4x)^-3/2 * (2x+4)  .  Rearranging and simplifying this expression gives:  dy/dx = -(x+2) / (x^2+4x)^3/2  .

So you've encountered something like: g(x) = | 4 f( |x + 1| - 3 ) + 5 | What a mess! And you are only given the graph of f(x). How do we go about sketching g(x)? As in any problem in mathematics, let's break it all down to a list of simple steps. We start with a graph of f(x). 1. Let's create a function f1(x) = f(x+1). You might already know that adding a number to the argument of a function is equivalent to 'sliding' its graph by that number to the left (it's quite simple, after all, you make an x=5 give f(5+1)=f(6) - which is to the right of f(5) we call that translation by a vector) 2. f1(x) is not quite like g(x) yet. Let's make f2(x) = f1( |x| ) = f( |x+1| ) . How does that affect the graph? As we know |x| = x, for positive x's and |x| = -x, for negative x's. So to sketch f2(x), we take the part of the graph of f1(x) which is to the right of the y-axis (where x is positive), and on the left half, we need to draw a mirror image of the right half. We're one step closer to g(x). 3. Next we need to include the number -3 in the argument. Let f3(x) = f2(x-3). Which means another shift along the x-axis. This time we move the graph of f2(x) to theright, by 'amount' 3.  4. Now we are 'outside' f3(x) - we are not going to play with the argument anymore. Which means that now we will change the graph in the y-axis only. The remaining steps include multiplying the function by 4. This means that everywhere on the graph, the function is 4 times farther away from the x-axis (from 0). So sketch f4(x) = 4*f3(x), 4 times taller and steeper. 5. Then there is the 5 left to add. If we add a number to the function, it means that it goes up in the y direction by 5, everywhere. So let's shift the graph upwards. f5(x) = f4(x) + 5 6. We are nearly done now. The last thing to do is the absolute value. Once again, for all positive values, it stays as is, and for all negative values it gets a minus sign. Thus everywhere the function f5(x) has values smaller than 0, we need to 'flip' it upside down (symmetry with respect to the x-axis). f6(x) = |f5(x)| g(x) = f6(x) And that is it. You now have an accurate depiction of g(x). It wasn't that complicated after all. You just needed to make incremental changes and step by step move 'from the inside out'. - See more at: https://www.mytutorweb.co.uk/tutors/secure/ta-yourexplanations.html#sthash.iwmqcICD.dpuf

The preparation for writing an essay is incredibly important in allowing you to structure your work in a way that is engaging, readable and above all, well set out, so that your own thoughts and analysis can shine through. It can seem daunting writing an essay plan, as this seems like more work on top of the essay itself, however a well thought out plan can save time and stress in the long run. If you are a visual person, perhaps make a mind map of all of the themes and topics in your chosen texts: connect links between different ideas and see where thoughts naturally go to. These will eventually make up your paragraphs. It can also be a good idea to format your essay by listing your paragraphs with their own individual sub-headers, starting with your 'Introduction' and finishing with your 'Conclusion'. Try and be strict with yourself and stick to your essay plan - this will help to avoid waffling and straying away from your core ideas and points of analysis. Remember to refer back to your essay plan whilst writing to cement your ideas and before you know it, even writing essays in exam conditions will seem easier because of past essay plan practice. 

For a general quadratic equation of the form ax² + bx + c = 0 (1) We will solve for x via the ‘Completing the Square‘ method:   First we divide equation (1) through by a on both sides, yielding: x² +(b/a)x + c/a = 0   Slightly rearranging, we then write (for tidiness): x² + (b/a)x = -c/a (2)   Now the key behind completing the square is to try and write the x² and x terms as the square of some quantity, which is close to:  (x + (b/2a))² (3)   What we have done is half the coefficient of the x term, as when the brackets are expanded out, you have a certain sum repeated twice, I will demonstrate in fuller detail now what I mean.   Expanding out (3), we have (x +(b/2a))(x+(b/2a)) = x²+ (b/2a)x + (b/2a)x + (b²/4a²) = x² + (b/a)x + (b²/4a²) , which is almost identical to the left hand side of equation (2), except now we have the extra third term which is (b²/4a²)   We can rearrange this easily to show that x²+ (b/a)x =(x +(b/2a))²-(b²/4a²) Substituting this identity into equation (2), we now have (x +(b/2a))² -(b²/4a²) = -c/a (4)   We now rearrange as follows: (x+(b/2a))² = (b²/4a²) - (c/a) = (b²-4ac)/(4a²) , where in the third equality I have simply summed over a common denominator by multiplying the second term by 4a over the numerator and denominator.   Finally we take the square root of both sides, and rearrange to arrive at the quadratic formula we were looking for, as follows: x = (-b ± sqrt(b²-4ac))/2a   Note that we have two solutions to the quadratic equation, resulting from the existence of both positive and negative square roots, hence the ± sign. Also ‘sqrt' denotes taking the square root.   Corollary   Note that the solution of x depends on the quantity sqrt(b2-4ac), which is known as the Discriminant. This is important as it results with 3 distinct cases for the solution which are as follows:   1) b² - 4ac > 0 , and so it gives one positive and one negative square root, leading to 2 unique solutions. 2) b² -4ac = 0, which then yields exactly one solution, known as a repeated root, -b/2a . 3) b² -4ac < 0, which cannot be square rooted within our common number system which we call the Real Numbers. If we extend our number system, to allow for square roots of negative numbers, we call these the Complex Numbers, which go far beyond the scope of the course.   If you want to find out more, be sure to stay on till A Level and take Further Maths!