What is the difference between these two equations, and their answers? I'm meant to find dy/dx, or just derivative of y.
$y_1= \cos(5x+1)^9$
&
$y_2= \cos^9(5x+1)$
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$y= \cos((5x+1)^9)$ is a composite of $f(x)=\cos x, g(x)=(5x+1)^9$ so by composition rule, the derivative is $y'=f'(g(x))g'(x)$
While $y= \cos^9(5x+1)$ is a composite of $f(x)=\cos^9 x, g(x)=(5x+1)$
John
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$$\begin{align} y_1 = \cos\Big((5x+1)^9\Big) \implies y_1' & = -\sin((5x+1)^9)\cdot (9(5x+1)^8)\cdot (5)\\\\ & = -45(5x+1)^8\sin((5x+1)^9)\end{align}$$
$$\begin{align}y_2 = \cos^9(5x+1) = \Big(\cos(5x+1)\Big)^9 \implies y_2' &= 9\cos^8(5x+1)\cdot (-\sin(5x+1))\cdot (5)\\ \\ &= -45 \sin(5x+1)\cos^8(5x+1)\end{align}$$
amWhy
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The second expression $\cos^9 (5x+1)$ is perfectly well-defined and means $(\cos(5x+1))^9$.
The first expression $\cos(5x+1)^9$ is somewhat ambiguous, since technically it could mean either $\left(\cos(5x+1)\right)^9$ or $\cos\left((5x+1)^9\right)$.
Having said this, however, I will quickly add that for trig functions, this would usually be interpreted as $\cos\left((5x+1)^9\right)$, since the form $\cos^9 (5x+1)$ would traditionally be preferred if the exponent were meant to be carried by the function value rather than the function argument.
This is analogous to the understanding that $\sin ab$ means $\sin (ab)$, not $(\sin a)(b)$; so also $\sin x^2$ means $\sin(x^2)$, not $(\sin x)^2$ (the latter would be written $\sin^2 x$).
To complicate things, expressions involving "ordinary" functions are usually interpreted differently; for example, $f(x)^2$ would mean $\left(f(x)\right)^2$, and not $f(x^2)$. I think the primary justification is that one usually omits the parentheses around simple arguments for trig functions, only supplying them when the argument has several terms. In the absence of parentheses, convention is used. So you usually will see $\sin x$ rather than $\sin(x)$, but $\sin(x+y)$ requires the parentheses because $\sin x + y$ means $(\sin x) + y$.
MPW
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