# Math Help - Derivative Help

1. ## Derivative Help

Hi, Im a total newb at Calculus, Im only in Triginometry in my school, but Im trying to self-learn Calculus, and im having trouble understanding this concept of derivatives. I understand that it is the tangent of a line, and I also understand how to find them the long way using, $\frac{f(x+h) - f(x)}{h}$

But my question is, is there any shortcut to finding a derivative, like in the function $f(x) = x^n$ the derivative is simply $f^1(x) = nx^n-1$, but what do I do for the bigger ones such as $f(x) = x^2 + 2x$, is there any shortcut there? I already found the derivative of $f(x) = x^2 + 2x$ which is $\frac{x^2 + 2x}{-4x}$

So please tell me if you know any shortcuts to finding the derivatives of a function such as $f(x) = x^2 + 2x$. Thanks!

P.S - Please dont use any big calculus words - im only in triginometry!

2. Originally Posted by gs.sh11
Hi, Im a total newb at Calculus, Im only in Triginometry in my school, but Im trying to self-learn Calculus, and im having trouble understanding this concept of derivatives. I understand that it is the tangent of a line, and I also understand how to find them the long way using, $\frac{f(x+h) - f(x)}{h}$

But my question is, is there any shortcut to finding a derivative, like in the function $f(x) = x^n$ the derivative is simply $f^1(x) = nx^n-1$, but what do I do for the bigger ones such as $f(x) = x^2 + 2x$, is there any shortcut there? I already found the derivative of $f(x) = x^2 + 2x$ which is $\frac{x^2 + 2x}{-4x}$

So please tell me if you know any shortcuts to finding the derivatives of a function such as $f(x) = x^2 + 2x$. Thanks!

P.S - Please dont use any big calculus words - im only in triginometry!
The derivative of $f(x) = x^2 + 2x$ is not what you said it was.

You said that you knew that $x^n$ differentiates to $nx^{n-1}$, and that's the only rule you need to know to differentiate the function you gave.

Because differentiation is a linear operator, you can break the function up into little parts, and differentiate them separately, and then add them back together at the end.

So first of all, find the derivative of $x^2$, which you know to be $2x$, then find the derivative of $2x$. Now, you can do that using the same rule because $x = x^1$. So the derivative of 2x is simply $2 \times (1 \times x ^{1 -1 }) = 2 \times x^0 = 2 \times 1 = 2$

So to conclude, the derivative of $x^2$ is $2x$, and the derivative of $2x$ is $2$, therefore, the derivative of $x^2 + 2x$ is $2x + 2$.

Doing it the 'long way', you should have this:

$\displaystyle f'(x) =\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) - x^2 - 2x}{h}$

$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 2x +2h - x^2 - 2x }{h} = \lim_{h \to 0} \frac{ 2xh+ h^2+2h }{h}$

$= \lim_{h \to 0} 2x +h+2 = 2x + 2$

3. Thankyou very much!! I understand this now!