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C4 Differentiation - 2 questions

Okay, the book says that if y=ax,then dydx=axlnay=a^x, \text{then }\frac{dy}{dx}=a^xlna. Then, in one of the questions I need to differentiate 2−x2^{-x}. So I use the general rule above and get dydx=2−xln2\frac{dy}{dx}=2^{-x}ln2, however they got dydx=2−xln2\frac{dy}{dx}=2^{-x}ln2. Am I missing something here?

Second question:

The book explains how to differentiate the general power function:
Unparseable latex formula:

\begin{array}{lcr}[br]y=a^x\\[br]lny=xlna\\[br]\therefore \frac{1}{y}\cdot \frac{dy}{dx}=lna[br]\end{array}



Ok.. stopping there, surely that's not correct? They differentiated lny, and by using the chain rule got 1y⋅dydx\frac{1}{y}\cdot \frac{dy}{dx}. Then, by definition, shouldn't you get lna+x⋅1a⋅dadxlna+x\cdot \frac{1}{a}\cdot \frac{da}{dx} for the RHS? How did they just get lna?
Original post by ViralRiver
Okay, the book says that if y=ax,then dydx=axlnay=a^x, \text{then }\frac{dy}{dx}=a^xlna. Then, in one of the questions I need to differentiate 2−x2^{-x}. So I use the general rule above and get dydx=2−xln2\frac{dy}{dx}=2^{-x}ln2, however they got dydx=2−xln2\frac{dy}{dx}=2^{-x}ln2. Am I missing something here?

The book is wrong and so are you.
Note that 2−x=(12)x2^{-x} = (\frac{1}{2})^x. Apply your rule to this and you will get dydx=−2−xln⁡2\frac{dy}{dx} = -2^{-x}\ln 2.

Second question:

The book explains how to differentiate the general power function:
Unparseable latex formula:

\begin{array}{lcr}[br]y=a^x\\[br]lny=xlna\\[br]\therefore \frac{1}{y}\cdot \frac{dy}{dx}=lna[br]\end{array}



Ok.. stopping there, surely that's not correct? They differentiated lny, and by using the chain rule got 1y⋅dydx\frac{1}{y}\cdot \frac{dy}{dx}. Then, by definition, shouldn't you get lna+x⋅1a⋅dadxlna+x\cdot \frac{1}{a}\cdot \frac{da}{dx} for the RHS? How did they just get lna?

That's correct. When you differentiate nxnx w.r.t. x, you get nn. (ln⁡a)x(\ln a)x is of this form.
I'm guessing you mean dydx=−2−xln2\frac{dy}{dx}= -2^{-x} ln 2. Remember to use the chain rule.

As for the second part, a is a constant, so its derivative is 0.
Reply 3
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ViralRiver
OP
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Original post by Farhan.Hanif93
The book is wrong and so are you.
Note that 2−x=(12)x2^{-x} = (\frac{1}{2})^x. Apply your rule to this and you will get dydx=−2−xln⁡2\frac{dy}{dx} = -2^{-x}\ln 2.


That's correct. When you differentiate nxnx w.r.t. x, you get nn. (ln⁡a)x(\ln a)x is of this form.


SOrry, I meant the book got -2^-xln2, which is what I don't understand. What do you mean by your response to the second question?
Reply 4
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Pheylan
similarly, what's ∫ex dx\displaystyle\int ex\ dx ?
Reply 5
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ViralRiver
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16
Original post by Pheylan
similarly, what's ∫ex dx\displaystyle\int ex\ dx ?


ex22+c\frac{ex^2}{2}+c
Original post by ViralRiver
SOrry, I meant the book got -2^-xln2, which is what I don't understand.

ddx[2−x]\frac{d}{dx}[2^{-x}]
=ddx[(12)x]= \frac{d}{dx}\left[(\frac{1}{2})^x\right]
=(12)xln⁡(12)=(\frac{1}{2})^x \ln (\frac{1}{2})
=2−xln⁡(2−1)=2^{-x}\ln (2^{-1})
=−2−xln⁡2=-2^{-x}\ln 2

What do you mean by your response to the second question?

a is not a variable, it's a constant. Therefore lna is a constant. When you differentiate nxnx for some constant coefficient, n, you get the derivative to be nn, right? C1 differentiation?
So ddx[xln⁡a]=ln⁡a\frac{d}{dx}[x\ln a] = \ln a.
Reply 7
Avatar for ViralRiver
ViralRiver
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Original post by Pork and Beans
I'm guessing you mean dydx=−2−xln2\frac{dy}{dx}= -2^{-x} ln 2. Remember to use the chain rule.

As for the second part, a is a constant, so its derivative is 0.


Well, how do we know a is a constant? I'm still unsure of how to get -2 etc for the first part, even with the chain rule >< - this doesn't seem to make any sense.
Reply 8
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ViralRiver
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Original post by Farhan.Hanif93
ddx[2−x]\frac{d}{dx}[2^{-x}]
=ddx[(12)x]= \frac{d}{dx}\left[(\frac{1}{2})^x\right]
=(12)xln⁡(12)=(\frac{1}{2})^x \ln (\frac{1}{2})
=2−xln⁡(2−1)=2^{-x}\ln (2^{-1})
=−2−xln⁡2=-2^{-x}\ln 2


a is not a variable, it's a constant. Therefore lna is a constant. When you differentiate nxnx for some constant coefficient, n, you get the derivative to be nn, right? C1 differentiation?
So ddx[xln⁡a]=ln⁡a\frac{d}{dx}[x\ln a] = \ln a.


Okay, I understand it a bit more now, but I still have a few problems. If y=2−xy = 2^{-x} then y=12xy = \frac{1}{2^x}, which doesn't make a difference in this question, but why did you raise the entire fraction (1/2) to the power of x?
Original post by ViralRiver
Okay, I understand it a bit more now, but I still have a few problems. If y=2−xy = 2^{-x} then y=12xy = \frac{1}{2^x}, which doesn't make a difference in this question, but why did you raise the entire fraction (1/2) to the power of x?

To apply the rule you stated. You can always do it the long way. To be in the form axa^x, which you want, you need the coefficient of x in the power to be 1. It's still equivalent because 1x=11^x = 1.
Reply 10
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ViralRiver
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Original post by Farhan.Hanif93
To apply the rule you stated. You can always do it the long way. To be in the form axa^x, which you want, you need the coefficient of x in the power to be 1. It's still equivalent because 1x=11^x = 1.


Perfect, thanks :smile: .

Is there a sort of general rule in a similar way to the following:?

d(sinf(x))dx=f′(x)cos(f(x))\frac{d(sinf(x))}{dx}=f'(x)cos(f(x))
Original post by ViralRiver
Perfect, thanks :smile: .

Is there a sort of general rule in a similar way to the following:?

d(sinf(x))dx=f′(x)cos(f(x))\frac{d(sinf(x))}{dx}=f'(x)cos(f(x))

Yes, that is true.
Reply 12
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ViralRiver
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Original post by Farhan.Hanif93
Yes, that is true.


Hmm, soryr I should be more clear.

I can use the chain rule to differentiate sin, as follows:

f(x)=h(g(x))f(x)=h(g(x))
f′(x)=g′(x)⋅h′(g(x)f'(x)=g'(x)\cdot h'(g(x)

then, if

f(x)=sin(x2),f′(x)=2xcos(x2)f(x) = sin(x^2), f'(x) = 2xcos(x^2)

but.. what if f(x)=axf(x) = a^x, how do you use the chain rule on that?

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