Hi!

How do you use calculus to solve the gradients of any tangent on any curve? Is there an equation, and if so, why does it work. Please consider the fact that I'm only 15 and don't even know what calculus is!!!!!!!!

Thanks.

d(nx^m + px^q )/dx = mnx^m-1 + pqx^q-1
This is the formula you need in this situation. It's called differentiation, and what that stuff meant was:

The gradient on the curve y = nx^m + px^q

is mnx^m-1 + pqx^q-1

And by the way, don't worry about asking a question that you think we'll laugh at; we're all grateful for a question we can answer, and we recognise that everyone on this site are different ages.

Here's a basic intro to calculus (for a fuller treatment, the book "Teach Yourself Calculus" was recommended to me by a member of this site, and it's a very good and comprehensive book):

Limits

a limit of an expression is the number that the expression goes to when the variable in the expression goes either very small or very large.

For example, the limit of

f(x)=^2x /_x+1

as x goes to infinity

is found by rewriting the expression as

f(x)=2/(1+[1/x])

And, as x-> infinity, 1/x vanishes to 0, telling us that the limit of the function is 2.

This is a very short intro, and once again, the book mentioned does a far better job explaining this than me.

Gradients

Now, if we let dx stand for the increase in x, and dy stand for the increase in y given a corresponding increase in x, we can see that (dy)/(dx) starts to get very close to the gradient equation of a tangent to a curve as dx gets very close to 0.

From slopes of lines, we also know that

if

y=f(x)

then

dy=f(x+dx)-f(x)

With this, we can start to find gradients of tangents. This process is called differentiation.

Differentiation

say that y=f(x)=x²

then using the earlier concepts in ''Gradients'',

dy=(x+dx)²-x²

=x²+2x×dx+(dx)²-x²

=2x×dx+(dx)²

Now, when this is divided by dx, and we let dx get very small, we have found the slope of a line tangent to the equation y=x²

Doing this

dy/dx=(2x×dx+(dx)²)/dx)

=2x+d

And, as dx gets very small, we can simply take it off the equation,

giving (if we let dx=dx as dx® 0)

dy/dx=2x

This should give you a general idea of how differentiation works. Once again, I would recommend that book to you as it is excellent and probably alot more understandable than this. On that note, if you don't understand some of this don't worry at all (after all, half of it literally is greek), -I had an incredibly hard time understanding these concepts when I first learned of them- simply post again, and I, or someone else I'm sure would be glad to answer your question. Hopefully that hasn't been too confusing, although it's bound to be a little (okay a lot) perplexing the first time you see that stuff.

Hope this helps,

Brad

I also forgot to mention that those sort of techniques can be used for all sorts of equations and functions.

Thanks. You are all right- I haven't a clue what you're going on about!!!!
Oliver, in your equation, what do p, q and n stand for? For example, how would I use the equation to find the gradient of a point on the graph y=x³ ?

Perhaps all you need to know is that the graph defined by y = xⁿ has a gradient given by the equation (dy/dx) = nx^n-1 for any number n. (dy/dx denotes 'gradient')

So for example, to find the gradient at the point defined by x=1 on the curve defined by y = x³ first of all we to find the 'gradient equation' i.e. dy/dx. That is dy/dx = 3x² . So at x=1 the gradient is 3 x 1² = 3.

The reason I included all those many variables, is because, in my opinion, it might not be obvious to someone who's never seen it before that you can differentiate x³ and 4x² separately when differentiating x³ + 4x² . p, q, etc. are the various co-efficients and indices of x, and can have any real value.