Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

Shuuya · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$\noindent We can do each of these (except the last one) using the fact that $\frac{\mathrm{d}}{\mathrm{d}x} \left(a^{mx+b}\right) = a^{mx+b}\cdot m \ln a$.$

$\noindent To prove this fact, we can let $y = a^{mx+b}$, and then we have $y=\mathrm{e}^{\left(mx+b\right) \ln a}$. Let $u=(mx+b)\ln a = m\ln a\cdot x + b\ln a$. Then $\frac{\mathrm{d}u}{\mathrm{d}x} = m\ln a$. Now by the chain rule, noting $y=\mathrm{e}^{u} = a^{mx+b}$, we have$

$\begin{align*} y^\prime &= \frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x} \\ &= \mathrm{e}^{u} \cdot \left(m\ln a\right) \\ &= a^{mx+b}\cdot \left(m\ln a\right), \end{align*}$

$\noindent as claimed.$

$\noindent The main rule is that $\frac{\mathrm{d}}{\mathrm{d}u} \left(a^u \right) = \ln a \cdot a^u$. This, along with the chain rule, allows us to do all the questions above.$

Thank you!

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$\noindent We can do each of these (except the last one) using the fact that $\frac{\mathrm{d}}{\mathrm{d}x} \left(a^{mx+b}\right) = a^{mx+b}\cdot m \ln a$.$

$\noindent To prove this fact, we can let $y = a^{mx+b}$, and then we have $y=\mathrm{e}^{\left(mx+b\right) \ln a}$. Let $u=(mx+b)\ln a = m\ln a\cdot x + b\ln a$. Then $\frac{\mathrm{d}u}{\mathrm{d}x} = m\ln a$. Now by the chain rule, noting $y=\mathrm{e}^{u} = a^{mx+b}$, we have$

$\begin{align*} y^\prime &= \frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x} \\ &= \mathrm{e}^{u} \cdot \left(m\ln a\right) \\ &= a^{mx+b}\cdot \left(m\ln a\right), \end{align*}$

$\noindent as claimed.$

$\noindent The main rule is that $\frac{\mathrm{d}}{\mathrm{d}u} \left(a^u \right) = \ln a \cdot a^u$. This, along with the chain rule, allows us to do all the questions above.$

Has there been many proof questions involving the chain rule in the past involving power functions in HSC exam papers? I've traditionally done it a slightly different way to your method.

$y=2^x$

$By taking natural logs of both sides ln y =x ln 2$

$$ By doing the derivative of both sides we get $ \frac{1}{y} \frac{dy}{dx} = ln 2$

$\frac{dy}{dx}=y \ ln 2$

$\frac{dy}{dx}= 2^x \ ln 2$

InteGrand · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Has there been many proof questions involving the chain rule in the past involving power functions in HSC exam papers? I've traditionally done it a slightly different way to your method.

$y=2^x$

$By taking natural logs of both sides ln y =x ln 2$

$$ By doing the derivative of both sides we get $ \frac{1}{y} \frac{dy}{dx} = ln 2$

$\frac{dy}{dx}=y \ ln 2$

$\frac{dy}{dx}= 2^x \ ln 2$

What you have done is known as logarithmic differentiation. I don't think 2U students are expected to know this (but it is definitely handy).

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
What you have done is known as logarithmic differentiation. I don't think 2U students are expected to know this (but it is definitely handy).

$$ If I were to do it Integrand's way, would it be something like this? $ \frac{d}{dx} (2^x)=e^{x ln 2}$

$u=x ln 2$

$\frac{du}{dx}=ln 2$

$$ Now, by the chain rule, noting $ y=e^u=2^x$

$$ We have $ \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$=e^u \ ln 2$

$= 2^x \ ln 2 $ as claimed . Correct me, if I am wrong $$

InteGrand · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
$$ If I were to do it Integrand's way, would it be something like this? $ \frac{d}{dx} (2^x)=e^{x ln 2}$

$u=x ln 2$

$\frac{du}{dx}=ln 2$

$$ Now, by the chain rule, noting $ y=e^u=2^x$

$$ We have $ \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$=e^u \ ln 2$

$= 2^x \ ln 2 $ as claimed . Correct me, if I am wrong $$

That is correct.

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I honestly didn't know until you, leehuan and Carrotsticks corrected me that you could have a log with base e, without the subscript being written. (Having doing 15 years of maths).

Always thought that a log without a base 'e' was to the base 10.

Shuuya · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
I honestly didn't know until you, leehuan and Carrotsticks corrected me that you could have a log with base e, without the subscript being written. (Having doing 15 years of maths).

Always thought that a log without a base 'e' was to the base 10.

I always get confused; do we assume that it is to the base 'e' if not specified, or base 10?

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$I've decided to submit Part 7 (b) to that question because I have nothing else to do on a Monday night$

$y=10^x$

$\frac{d}{dx}(10^x)=e^{x ln 10}$

$u=x ln 10$

$\frac{du}{dx}= ln 10$

$Now, by the chain rule, noting y=e^u =10^x$

$$ We have $ \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$= e^u \ ln 10$

$= 10^x \ ln 10$

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Shuuya said:
I always get confused; do we assume that it is to the base 'e' if not specified, or base 10?

$I've always thought that $ y=log_{e}x= ln x $ are the same thing. The Cambridge book answers write their answers as y=log x . I think if in doubt go for the natural log.$

InteGrand · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Shuuya said:
I always get confused; do we assume that it is to the base 'e' if not specified, or base 10?

Base e.

Shuuya · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
$I've always thought that $ y=log_{e}x= ln x $ are the same thing. The Cambridge book answers write their answers as y=log x . I think if in doubt go for the natural log.$

Yeah, that's what I thought too. In Fitz Patrick, I think the answers refer to y=log(x) as base 10 so its annoying haha

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$ Part 7 (c) $ y=\pi^x$

$\frac{d}{dx} (\pi^x)= e^{x \pi}$

$u=x \ ln \pi$

$\frac{du}{dx} = ln \pi$

$$ Now by the chain rule, noting y=e^u=\pi^x$

$$ We have $ \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$=e^u ln \pi$

$= \pi^x ln \pi$

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$ Part 7 (d) $ y=a^x$

$\frac{d}{dx} (a^x)= e^{x a}$

$u=x \ ln \ a$

$\frac{du}{dx} = ln \ a$

$Now by the chain rule, noting y=e^u=a^x$

$$ We have $ \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$=e^u ln a$

$= a^x ln \ a$

Shuuya · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
$$ Part 7 (d) $ y=a^x$

$\frac{d}{dx} (a^x)= e^{x a}$

$u=x \ ln \ a$

$\frac{du}{dx} = ln \ a$

$Now by the chain rule, noting y=e^u=a^x$

$$ We have $ \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$=e^u ln a$

$= a^x ln \ a$

Yay you happened to do all three of the questions that I needed to do for homework, thanks

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I'm just bored I want to do all of them for a giggle.

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$ Part 7 (e) $ , y=2^{3x-1}$

$y=2^{3x-1}$

$y=3^{(3x-1) ln 2}$

$u=(3x-1) ln 2$

$\frac{du}{dx} = 3 ln 2$

$$ Now by the Chain rule, noting y=e^u=2^{3x-1}$

$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$= e^u \ 3 ln 2$

$= 2^{3x-1} \ 3 ln 2$

$= \ 2^{3x-1} ln 8$

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$ Part 7 (f) $ , y=5^{2-x}$

$y=e^{(2-x)ln 5}$

$u=(2-x) ln 5$

$u= 2 ln 5- x ln 5$

$\frac{du}{dx} = - ln 5$

$$ Now by the Chain rule, noting y=e^u=5^{2-x}$

$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$= e^u ( - ln 5)$

$=-5^{2-x} ln 5$

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$ Part 7 (g) $ , y=a^{x-2}$

$y=e^{(x-2)ln a}$

$u=(x-2) ln a$

$u=x ln a-2 ln a$

$\frac{du}{dx} =ln a$

$$ Now by the Chain rule, noting $ y=e^u=a^{(x-2)}$

$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$= e^u \ a^{x-2}$

$= a^{x-2} \ ln a$

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$ Part 7 (h) $ , y=a^{bx+c}$

$y=e^{(bx+c)ln a}$

$u=(bx+c) ln a$

$u=bx \ ln a+c \ ln a$

$\frac{du}{dx} =b \ ln a$

$$ Now by the Chain rule, noting $ y=e^u=a^{(bx+c)}$

$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$= e^u \ b \ ln a$

$= a^{bx+c} \ b \ ln a$

davidgoes4wce · Feb 22, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$ Part 7 (j) $ , y=3^{x^3-3x}$

$y=e^{(x^3-3x)ln 3}$

$u=(x^3-3x) ln 3$

$u=x^3 ln 3-3 x ln 3$

$\frac{du}{dx} = 3 x^2 ln 3-3 ln 3$

$$ Now by the Chain rule, noting y=e^u=3^{x^3-3x}$

$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$

$= 3^{x^3-3x} \ 3 ln 3 (x^2-1)$

Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

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