# Math Help - Help! functions and derivatives

1. ## Help! functions and derivatives

Hi,

Q1.
For f(x) = 2x^3 + 5x^2 - 7x -4 find f '(x) using the limit definition of the derivative.

Q2.
Find the derivatives of f(x) = cos(4x) + sin(5x) up to the fourth order, f '(x), f ''(x), f '''(x) and f ''''(x).

Confused on these questions - i know derivatives, but not that advanced,

thanks

Moon

2. I haven't done limits in a long time, so I'm not sure on how to go about solving the first one.

However, for f(x) = cos(4x) + sin(5x) the derivatives are not too hard. As sin and trig are periodic functions, the derivatives will alternate in a pattern.
For example, if f(x) = cosx, f'(x)=-sinx, f''(x)=-cosx, f'''=sinx, f''''=cosx.

So he derivatives would be:
f'(x) = -4sin(4x) + 5cos(5x)
f''(x) = -16cos(4x) - 5sin(5x)
f'''(x) = 64sin(4x) - 25cos(5x)

Then can you figure out the fourth one on your own?

3. Originally Posted by Moon Hoplite
Hi,

Q1.
For f(x) = 2x^3 + 5x^2 - 7x -4 find f '(x) using the limit definition of the derivative.

Q2.
Find the derivatives of f(x) = cos(4x) + sin(5x) up to the fourth order, f '(x), f ''(x), f '''(x) and f ''''(x).

Confused on these questions - i know derivatives, but not that advanced,

thanks

Moon
Q1)
This one is all about doing the algebra.

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
$=\lim_{h\rightarrow 0}\frac{[2(x+h)^3 + 5(x+h)^2 - 7(x+h) -4]-[2x^3 + 5x^2 - 7x -4]}{h}$
...

Q2)
Remember:
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}[f(g(x))] =\frac{d}{dx}f(g(x))\cdot \frac{d}{dx}g(x)$ (Chain Rule)

I will start you off:
$f(x) = \cos(4x) + \sin(5x)$
$f'(x) = -4sin(4x) + 5\cos(5x)$
$f''(x) = \frac{d}{dx}(f'(x))= -16\cos(4x) - 25\sin(5x)$
...

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Edit: xxlvh beats me to it.