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I am having some trouble with this question, I was hoping you could help me.

Find the derivatives of the following:

f(x)=sin(2x)/(3x^2)

and

f(x)=5e^(2x)cos(3x^2)

After having done so, are the resulting derivatives equal to eachother?

Thanks,

Jenny

To find a derivative of a quotient, the rule is if f(x) = g(x)/h(x),

the derivative is defined, as I remember it, as "low d high - high d lo all over lo squared.

Here, g(x) = high and h(x) = low.

That said, f'(x) = (h(x)g'(x) - g(x)h'(x))/h²(x).

In the first case, we have g(x) = sin(2x) and h(x) = 3x².

From here, we can see that g'(x) = 2*cos(2x) and h'(x) = 6x.

This gives f'(x) = [(3x²)(2*cos(2x)) - sin(2x)(6x)]/9x^4.

That can be rewritten as f'(x) = 6x[x*cos(2x) - sin(2x)]/9x^4.

Reducing gives f'(x) = 2[x*cos(2x) - sin(2x)]/3x³.

For f(x) = 5(e^x)cos(3x³), let g(x) = e^x and h(x) = cos(3x³).

From here, it can be seen that g'(x) = e^x and h'(x) = (-9x²)*sin(3x³).

Putting this all together gives us f'(x) = 5[cos(3x³)e^x - (e^x)(-9x²)*sin(3x³)/cos²(3x³).

If we divide each term by cos(3x²), we get f'(x) = 5[e^x - (e^x)(-9x²)*tan(3x³)/cos(3x³).

It can also be seen a e^x can be factored out, giving

f'(x) = 5(e^x)[1 + 9x²*tan(3x³)]/cos(3x³).

As can be seen, these functions are not equal to each other. This can be realized from the start since one function contains an e^x and the other function doesn't.

Calculus

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