Differentiating Powers of x

Picture a skate ramp that starts almost flat and curls up into a near vertical wall. Walk on at the low end and it is an easy stroll. Ten steps later, the same ramp is a climb. One ramp, and the steepness is different everywhere you stand. A straight staircase is nothing like that. Every step rises by the same amount, so one number describes the whole flight. A curve has no single steepness. Ask how steep this curve is, and the question is unfinished until you say where. Differentiation finishes it: hand it the equation of a curve, and it hands back a formula for the steepness at any point.

Start with what gradient means on something straight, because that is what makes a curve strange. One note on where this lives. Differentiation sits on the Edexcel International GCSE Higher papers and the Cambridge IGCSE extended papers, not on the AQA, OCR, Eduqas or Edexcel UK GCSE courses. Take the line y equals two x plus three. Step one across and you rise two, so the gradient is two. Gradient is rise divided by run, and on a straight line that ratio is the same at every point. Now the curve y equals x squared. Near the origin it is nearly flat. Out at x equals three it is climbing hard. Same equation, completely different steepness. The steepness at a point on a curve means the steepness of the straight line that just touches it there. Three points, three touching lines, tilted three different ways. You can draw that line and measure it by hand, but that method belongs to Estimating the Gradient of a Curve, in the chapter on real-world graphs. Differentiation skips the drawing. Feed it the equation of the curve and it gives back a second expression, a gradient formula, and that one formula covers every point on the curve at once. That formula is written d y by d x. It is one symbol, not d times y over d times x, and it means the rate at which y changes as x changes. Quick check before any method. Two equations: y equals two x plus three, and y equals x squared. Which one has the same gradient at every point? Pick one. I'll wait. The answer is y equals two x plus three. A straight line has gradient two everywhere. The curve has a different gradient at every single point, and that is the situation differentiation was built for. Everything from here is the machinery for producing that gradient formula.

Here is the machine itself, one rule doing two jobs in the same move. A term looks like a x to the power n: a number in front, x raised to a power. Differentiating it does two things. Multiply the term by the power, then reduce that power by one. Say it as a chant while your pen moves. Bring it down, knock it down. Bring the power down to the front and multiply, then knock the power itself down by one. Try it on x to the power five. Bring the five down, knock the power to four, and the gradient formula is five x to the power four. Now three x squared. Bring the two down and multiply by the three already there, giving six. Knock the power to one. Three x squared differentiates to six x. The rule is not arbitrary. On y equals x squared, step a tiny distance h to the right: the height rises by two x h plus h squared, over a run of h, so that step has gradient two x plus h. Shrink h to nothing and the gradient is exactly two x, which is what the rule gives you. Your turn on a single term. Differentiate four x cubed. Is the answer twelve x squared, or four x squared, or twelve x cubed? Take your pick. I'll wait. The answer is twelve x squared. Bring the three down and multiply by the four, giving twelve. Knock the power to two. Four x cubed differentiates to twelve x squared. If you picked four x squared, you knocked the power down but never brought it down and multiplied. Both halves of the chant happen, on every term. One term at a time is the whole skill, and the next step is doing it more than once.

Real questions hand you several terms at once, so here is how the rule scales up. Differentiate y equals three x to the power four, plus two x squared, minus seven x. Take the terms one at a time and never let them mix. First, three x to the power four. Bring the four down: four times three is twelve. Knock the power to three. That term gives twelve x cubed. Second, two x squared. Bring the two down: two times two is four. Knock the power to one. That term gives four x, because x to the power one is just x. Third, minus seven x. This one hides its power, because x on its own means x to the power one. Bring that one down, and multiplying by one changes nothing. Knock the power to zero, and anything to the power zero is one, so the term is just minus seven. The minus travels with the term the whole way. It does not get dropped, and it does not quietly become a plus. Line the three results up in the order they arrived. d y by d x equals twelve x cubed, plus four x, minus seven. One more term type, because real questions use it constantly. A plain number on the end, like plus nine, differentiates to zero. Adding nine lifts the whole curve nine units up the page without tilting it anywhere. The steepness at every point is untouched, so a term that changes no gradient adds nothing to the gradient formula. Your turn with the full method now. Differentiate five x cubed, minus two x squared, plus x. Pause it there and work it out. I'll wait. The answer is fifteen x squared, minus four x, plus one. Three times five is fifteen, power dropping to two. Two times two is four, staying negative, power dropping to one. And a plain x gives one. Term by term, sign by sign, and the mechanics are finished.

Knowing the rule is only half the job. The other half is spotting that a question wants it. On the harder questions nobody writes the word differentiate for you. The instruction arrives dressed as something else, and the first mark sits behind you recognising it. Three phrasings mean differentiate. Find the rate of change. Find d y by d x. Find the gradient of the curve. Those three are one instruction in three coats. Each asks how fast something is changing, not how much of it there is. Here is one of them in the wild. The height of a ball, in metres, t seconds after being thrown, is modelled by h equals twenty t, minus five t squared, plus one. Find an expression for the rate at which the height is changing, d h by d t. Have a think. I'll wait. The answer is d h by d t equals twenty minus ten t. Rate of change means differentiate. Twenty t gives twenty. Minus five t squared gives minus ten t. The plus one is a plain number, so it goes to zero. The letters changed and nothing else did. Height h against time t, so the formula is written d h by d t. The two moves never care what the letters are called. Then stop. That expression is the finished answer. Putting a number into d h by d t, for the rate at one particular moment, is the next video's job rather than this one's.

Now the examiners' own words, from the report on a twenty twenty four Edexcel International GCSE Higher paper, looking back over that paper as a whole. Students were less successful in producing a full treatment for surds, question seventeen, differentiating functions, question eighteen, and completing the square, particularly for negative quadratics, question twenty five. Read the phrase carrying the weight: a full treatment. Not started, not the first term right, but every term, signs intact. That is why each term here got said out loud on its own. Speed on this topic comes from being slow with each term. The second line from that report is about recognition, on a question where a curve's equation was given and calculus was the way in. Students who appreciated that this question required calculus usually gained at least two marks by differentiating y equals x cubed minus forty x plus one correctly. Many students correctly equated their d y by d x to eight to gain the first three marks. Appreciated that this question required calculus. That clause is the hinge. The marks only start once a student sees which technique the question wants, and the differentiating alone accounted for two of them. That last sentence goes one step past this video. Equating d y by d x to a number is the next topic, and it only earns anything if the differentiating underneath is right. Recognise it, differentiate term by term, then stop when the expression is written down.

Right, everything in one place. A straight line carries one gradient end to end. A curve carries a different one at every point, and differentiation produces the formula that gives it to you. For a single term, the chant does the work. Bring it down, knock it down. Multiply the term by its power, then reduce that power by one. A sum goes term by term, in order, carrying every sign. A plain number differentiates to zero, because sliding a curve up the page never changes how steep it is. And three phrasings tell you to reach for it: find the rate of change, find d y by d x, find the gradient of the curve.

Next in the chapter: Stationary Points, Maxima and Minima, where that gradient formula gets put to work.

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