I am attempting to find the derivative of $\log_{10} (x^3 + 1)$ I am not too sure to do with this actually, I know the formula states that it should be $1/ (x\ln a)$ but does that mean just plug it in to that or do I need to use the product rule? I am not too sure what to do.
Finding the derivatives of logarithms
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calculus
2 Answers
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The formula for the derivative of a logarithm says $$\frac{d}{du} \log_a(u) = \frac{1}{u\ln(a)}.$$ What you have here is $\log_{10}(x^3+1)$, so you need to use the Chain Rule: $$\frac{d}{dx}\log_{10}(x^3+1) = \frac{1}{(x^3+1)\ln(10)}\left(\frac{d}{dx}(x^3+1)\right).$$
There are no products in your function, so the Product Rule is not "in play".
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0...and since it'll come up often, one might want to remember the pattern $\dfrac{f'(x)}{f(x)}$... – 2011-10-01
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0I am a little confused then, why am I using the chain rule if x is just $x^3+1$ why am I using the chain rule also? – 2011-10-01
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0@Jordan: You are using the Chain Rule because this is a composition: it's the composition of the function $f(x)=x^3+1$, and the function $g(u) = \log_{10}(u)$. You *always* use the Chain Rule when you have to do the derivative of a composition. That's what it's for! Will you be less confused if I change the $x$ to a $u$ in the formula for the derivative of the logarithm? – 2011-10-01
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0Okay, I was just a little confused by the $1/xlna$ I guess I can't really just use that alone can I? It would only work if it was $log_a(x)$ or something similar, where x is just a single variable or constant? – 2011-10-01
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0@Jordan: It works *just like* all the other differentiation formulas. $\frac{d}{dx}\sin(x) =\cos(x)$ only works as-is if the thing inside the sine, $x$, is the variable with respect to which you are differentiating (the "$dx$" at the bottom of the Leibnitz operator $\frac{d}{dx}$). $\frac{d}{dx}e^x = e^x$ only works as-is if the exponent is exactly the same thing with respect to which you are taking the derivative. $\frac{d}{dx}x^3 = 3x^2$ only works if the thing you are cubing, $x$, is exactly the same thing with respect to which you are taking the derivative. (cont) – 2011-10-01
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0@Jordan: If it is anything else, you need to use the Chain Rule: $$\frac{d}{dx}\sin(u) = \cos(u)\frac{du}{dx}.$$ $$\frac{d}{dx}e^{u} = e^u\frac{du}{dx}.$$ $$\frac{d}{dx} u^3 = 3u^2\frac{du}{dx}.$$And the same is true for the logarithm. $\frac{d}{dx}\log_a(x) = \frac{1}{x\ln(a)}$ only works as-is if the thing in the logarithm, $x$, is exactly with respect to what you are differentiating. Otherwise, you need to use the Chain Rule:$$\frac{d}{dx}\log_a(u) = \frac{1}{u\ln(a)}\frac{du}{dx}.$$ – 2011-10-01
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0Ok, one thing then if I am doing implicit differentiation and I trying to find the derivative of siny+six then I would use the chain rule for siny and it would be cosy(yprime)? – 2011-10-01
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0@Jordan: You use the Chain Rule exactly like I wrote above:$$\frac{d}{dx}\sin(y + 6)= \cos(y+6)\frac{d}{dx}(y+6) = \cos(y+6)(y' + 0) = y'\cos(y+6).$$If you meant $\sin(y) + 6$, then $$\frac{d}{dx}(\sin y + 6) = \frac{d}{dx}\sin y + \frac{d}{dx}6 = \cos y\frac{dy}{dx} + 0 = y'\cos y.$$Either way, the answer is *not* $\cos(y')$, because you *never* take the derivative of the "outside function" and the "inside function" (in this case, sine and $y$, respectively) in the same step. That's supposed to get drilled into you when you learn the Chain Rule, though far too many students miss it. – 2011-10-01
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0I understand that I just don't quite understand why it is different than finding the derivative of sinx, that is cosx, I know this because it is on a table I had to memorize. Except when I get the derivatve of (I don't know the other notation so this won't make sense) cosy with respect to x it is cosy(yprime)? – 2011-10-01
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0@Jordan: It's not like differentiating $\sin x$, because in $\sin x$ you are differentiating **with respect to $x$**. The change in the varible is just *one step* removed from the function. If you differentiate $\sin y$ **with respect to $x$**, then *first* $y$ changes when $x$ changes, and *then* $\sin y$ changes because $y$ changed. So you need to take into account *both* how $y$ changes, and how $x$ changes. An analogy follows in a separate comment: – 2011-10-01
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0@Jordan: Suppose that you get 10% of everything I earn. If my income changes, then *your* income changes by 10% of whatever my income changes. You are just one step removed from me, so if we see how *my* income changing affects you, you get it directly by just multiplying your "cut" by my change. But suppose that *my* income is obtained as a 15% cut of what my neighbor makes. If you want to know how *your* income changes relative to how **my neighbor's income** changes, then first you need to go through *my* change (15%), and then how *my* change affects you (10% of 15% = 1.5 %) – 2011-10-01
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1@Jordan: So *your* rate of change relative to *my* income is 0.1 (10% of whatever my change is). **My** rate of change relative to my neighbor's income is 0.15 (15% of whatever his change is). But **your** rate of change relative to **my neighbor's income** is not 10%, and it's not 15%, it's the product of your change realtive to me, times my change relative to him: $.1\times .15 = .015$, or 1.5%. You don't use the same formula to compute both, because you are trying to figure out the change *with respect to different things*. – 2011-10-01
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0Okay that makes sense, so since we don't know what y is, and we are getting a derivative in respect to x and we know y changes as x changes we treat it like a function. Does that sounds about right? – 2011-10-01
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0@Jordan: Since $y$ changes as $x$ changes, we treat $y$ as a function of $x$. Since we don't know what $y$ is, we have to "leave it indicated" as $y'$; otherwise, I think you are on the right track. – 2011-10-01
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Without using chain-rule we can do in this way.
Let $log_{10}(x^3+1) = y \implies x^3+1= 10^y$
Now differentiating on both sides , $3x^2 = (ln10)(10^y)\left(\frac{\displaystyle dy}{\displaystyle dx}\right)$
So $\frac{\displaystyle dy}{\displaystyle dx}= \frac{\displaystyle 3x^2}{\displaystyle (x^3+1)(ln10)}$
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1Hate to break it to you, but you *are* using the Chain Rule: you use it when you claim that the derivative of $10^y$ is $(\ln 10)10^y\frac{dy}{dx}$; that last bit? That's the Chain Rule. – 2011-10-02