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$$\frac{d}{dx}(\cos^6x)$$

Using the chain rule $ M'(N(x)).N'(x)$, I'm deconstructing the $\cos$ function

$$\begin{align*} &M= \cos^6 \\ &N= x\end{align*}$$

End result should be $$-6\sin^5x \cdot 1$$ or $$-6\sin^5x$$

Yet my book said the end result is $$-6\sin x \cos^ 5 x$$

Why?

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    $\begingroup$ The chain rule. Let $f(y)=y^6,g(x)=\cos x$. Then your expression is $f(g(x))$. So the derivative is $f'(g(x))\cdot g'(x)=(6\cos^5x)(-\sin x)$. $\endgroup$ – almagest May 26 '16 at 8:04
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    $\begingroup$ You got it wrong fella, think about $\cos x$ and $x^6$ $\endgroup$ – windircurse May 26 '16 at 8:04
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    $\begingroup$ You have $f(x) = cos^6 x$ and you need to find f'(x). You used chain rule and now you have $M(x) = cos^6 x$ and you need to find M'(x)... Initial problem have reduced to itself ($f \equiv M$) $\endgroup$ – Pavel Mayorov May 26 '16 at 8:11
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    $\begingroup$ @AndyK He means that, if you write (as you suggested) $\cos^6x=f\circ g(x)$ with $f(x)=\cos^6x$ and $g(x)=x$ you, can of course, use the chain rule... in an astoundingly uneffective way: $$(\cos^6)'(x)=(f\circ g)'(x)=f'(g(x))\cdot g'(x)=(\cos^6)'(x)\cdot 1$$ By the way, there is a typo in your formulation of the chain rule. $\endgroup$ – user228113 May 26 '16 at 8:13
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    $\begingroup$ $M'(N(x))\cdot N'(x)$. $\endgroup$ – user228113 May 26 '16 at 8:24
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It's not true that $\frac{d}{du} \cos^6 u = -6 \sin^5 u$.

Hint Write $\cos^6 x$ as the composition $(f \circ g)(x)$, where $$f(u) = u^6 \qquad \textrm{and} \qquad g(x) = \cos x$$ (and then apply the chain rule).

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    $\begingroup$ You're welcome. $\endgroup$ – Travis Willse May 26 '16 at 8:06
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Actually I have found that the concept of chain rule which deals with derivatives of composition of functions is best explained in terms of what one might call "inner" and "outer" function. If we are considering the composition $$(f\circ g)(x) = f(g(x))$$ then $g$ is the inner function and $f$ the outer function. Or perhaps we can think of the order in which the functions are applied on $x$ to get the value of $(f\circ g)(x)$. The one which comes first in order is the inner function and the one which comes after is the outer function.

Now the chain rule says that to differentiate a composite functions the derivatives have to be taken in reverse order of composition so that first we differentiate the outer function (its argument is the same after differentiation) and then multiply it with the derivative of the inner function.

Let's take the case of $f(x) = \cos^{6}x$. This is clearly a composite function and to find the inner/outer functions here one should ask the following question: Starting from $x$ how do I get to the value $\cos^{6}x = f(x)$? The answer is simple: first I apply $\cos$ on $x$ to get $\cos x$ and next I raise the result to $6^{\text{th}}$ power. So $\cos$ is the inner function and raising to power $6$ is the outer function.

To differentiate $f(x)$ we perform differentiation in reverse order. Thus we first differentiate the function which raises $\cos x$ to $6^{\text{th}}$ power. Raising to a fixed power reminds us of $x^{n}$ and its derivative $nx^{n - 1}$ and here $n = 6$. So we have the term $6(\cos x)^{5}$ and then multiply this by the derivative of inner function $\cos$ i.e. we multiply it by $-\sin x$. The final answer is thus $f'(x) = \{6\cos^{5}x\}(-\sin x) = -6\sin x \cos^{5}x$.


I think it is better to take one more example related to the above function $f(x)$. Let us consider $g(x) = \cos (x^{6})$ which is also written as $g(x) = \cos x^{6}$. We again ask the question: how do I get from $x$ to $\cos x^{6}$? Well I now raise $x$ to $6^{\text{th}}$ power first and apply $\cos$ on the result.

To differentiate I proceed in reverse order. I first differentiate $\cos$ to get $-\sin (x^{6}) = -\sin x^{6}$ and multiply it with the derivative of $x^{6}$ i.e. $6x^{5}$ and thus $g'(x) = (-\sin x^{6})(6x^{5}) = -6x^{5}\sin x^{6}$.


The hurdle in applying chain rule comes because of difficulty in identifying the inner/outer functions. Note that these functions must themselves be simple enough (in the sense that they are not themselves composite) and there should be a direct formula available to differentiate them. The mistake by OP here is to identify the functions as $\cos^{6}$ and $x$. This is wrong because $\cos^{6}$ is the function which is itself composite of $\cos$ and raising to power $6$. You always need to pick the functions which are not composite themselves.

Naturally the chain rule applies to a composition of more than two functions also. And again we need to find exactly how we can reach from $x$ to the final value of function for $x$ and each step must be a simple/non-composite function applied to the result of previous step. Differentiating the final composite function would then work by taking derivatives of functions in reverse order i.e. the function of last step is differentiated first. Thus the function $$F(x) = \cos\log\arctan x^{2}$$ is thought of in steps as $$x \to x^{2} = A \to \arctan(A) = B \to \log B = C \to \cos C = F(x) \text{ finally!}$$ and then by reversing the steps $$F'(x) = (-\sin C)\left(\frac{1}{B}\right)\left(\frac{1}{1 + A^{2}}\right)(2x)$$ or $$F'(x) = -\sin(\log\arctan x^{2})\cdot\frac{1}{\arctan x^{2}}\cdot\frac{1}{1 + x^{4}}\cdot 2x$$ which can be simplified further.

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    $\begingroup$ Hi @paramand-singh, many thanks for your answer. Your pedagogy is sound but you can gain improvement by shortening the number of steps and making your explanation more visual. You try a sequential approach, which is good however, through a something more visual (thanks MTV for all the visual you brought -_-) , you can gain more people who will come out with a Oh I see. I wish that I still have that website's url I used to explain factorization to my GF's son, which was just insanely simple and mind blowing. Anyway, my 2 cents ;) $\endgroup$ – Andy K May 26 '16 at 16:16
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    $\begingroup$ Thanks @AndyK. I fully agree with you comment.I will perhaps create a visual image using some tool later. $\endgroup$ – Paramanand Singh May 26 '16 at 16:28
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I did not want to put an answer but it is always good to have a step-by-step approach to things. It helps you out for your own understanding and your efforts can be used by others.

$$\frac{d}{dx}(\cos^6x)$$

You have to see $$\cos^6x$$ as $$(cosx)^6$$

Or as Travis said $f(u)=u^6$.

Once understood, resolution is or at least should be, easy.

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  • $\begingroup$ I wish to have a slow motion gif for that $\endgroup$ – Andy K May 26 '16 at 10:55

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