Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I differentiate $5^{x \cos x}$? From my book, it should be implicit differentiation, but how do I start?

If I let $u = x \cos x$, then I get

$$ \begin{eqnarray} \frac{dY}{dX} &=& u \cdot 5^{u - 1} \cdot \frac{du}{dX} \\ &=& x \cdot \cos x \cdot 5^{x \cos x-1} \cdot (\cos x-x \sin x) \end{eqnarray} $$

I don't suppose I did it right... The correct answer is

$$ \ln5 \cdot 5^{x \cos x}\cdot (\cos x-x \sin x) .$$

share|improve this question
1  
We have $5=e^{\ln 5}$. So our function is $e^{(\ln 5)(x\cos x)}$. Now use the Chain Rule. Or else let $y=5^{x\ln x}$. Then $\ln y=(x\ln x)(\ln 5)$. Differentiate both sides with respect to $x$, using implicit differentiation for the left-hand side. –  André Nicolas Oct 17 '11 at 6:34
    
What happens if you take the logarithm? –  Mark Bennet Oct 17 '11 at 6:35
    
Ah ... 1 thing I dont get is how is $5=e^{ln5}$. Also how did you convert $xcosx \rightarrow x ln x$ –  Jiew Meng Oct 17 '11 at 6:38
1  
@jiewmang: The logarithm is the inverse of the exponential function, so $\exp\ln x=\ln\exp x=x$. I think André made an error when he spoke of $x\ln x$, I think he meant $x\cos x$. –  anon Oct 17 '11 at 6:43
    
Oh I think I am getting you, the part about inverse ... but what bothers me is I could explain the 2nd part, $ln(e^x)=xlog_e e=x$, but I can't derive x from $e^{lnx}$. Apart from "understanding" $\ln{x}$ and $e^x$ are inverse –  Jiew Meng Oct 17 '11 at 7:13

1 Answer 1

up vote 1 down vote accepted

Your error is that the derivative of $5^u$ (with respect to $u$) is not $u5^{u-1}$; this isn't $u$ to a power but rather a base to the power of $u$. The usual route is with André's advice: write the function as an exponential, so $5^{x\cos x}=(e^{\ln5})^{x\cos x}=\exp(\ln5\cdot x\cos x)$. Here we use the fact that the natural logarithm and exponential functions are inverse functions of each other, so $\exp\ln a=\ln\exp a=a $.

Now apply the chain rule, and remember the exponential function is its own derivative.

share|improve this answer
    
I wonder why no one mentioned the rule:$\frac{d}{dx}a^{x}=a^x.\ln a$ for $a>0$. –  Tapu Oct 17 '11 at 10:57
    
@Swapan: I personally find it more convenient to remember that the natural exponential function is its own derivative, and then use the chain-rule subsequently... –  J. M. Oct 17 '11 at 11:00
    
@Swapan: Because the rule is derived first by making $a^x$ into $\exp(x \ln a)$ and then applying the chain rule, as here. When learning theory one shouldn't use rules before one knows why they work... –  anon Oct 17 '11 at 11:00
    
@J.M. and anon: My point was just a mention...may be then how it works or in reverse order. Once, simplified (and probably knowing how it works), I always found it is useful to use the ready-made rule;, e.g., to use Binomial Theorem to expand $(a+b)^8$ than multiplying 8 times or in this case using the original definition of derivative as limits :) –  Tapu Oct 17 '11 at 11:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.