Author Topic: Differentiation... (Read 2653 times) Tweet Share

squance · « **on:** April 08, 2008, 09:52:30 pm »

Can someone please help me out with some differentiation???

I keep getting the wrong answers....

Find the derivative of the following:

y(x) = x/(x^2+1)^1/2

I got (x^2+1) - 2x/(x^2+1)^3/2 as the answer to the first one... but the answer is 1/(x^2+1)^3/2

h(x) = -3x/2(1-x^2)^1/4

Im screwed for the second one.,.....
the answer for the second one is 3/4(1-x^2)^-5/4(x^2-2)

I know how to do these but its messy but i keep getting them wrong.. we use quotient rule for both...I need a solution please.

ed_saifa · « **Reply #1 on:** April 08, 2008, 09:57:19 pm »

Perhaps make it a negative power and then use the product rule?

Mao · « **Reply #2 on:** April 08, 2008, 10:01:32 pm »

$y=\frac{x}{\left(x^2+1\right)^{\frac{1}{2}}}$

using the quotient rule $y'=\frac{u'v-uv'}{v^2}$

$\frac{dy}{dx}=\frac{\left(x^2+1\right)^{\frac{1}{2}}-x^2\left(x^2+1\right)^{-\frac{1}{2}}}{x^2+1}$

multiplying by $\left(x^2+1\right)^{\frac{1}{2}}$ on top and bottom:

$\frac{dy}{dx}=\frac{\left(x^2+1\right)-x^2}{\left(x^2+1\right)^\frac{3}{2}}$

$\frac{dy}{dx}=\frac{1}{\left(x^2+1\right)^\frac{3}{2}}$

second question:

$h(x) = -\frac{3x}{2(1-x^2)^\frac{1}{4}}$

$h(x)=-\frac{3}{2}\cdot \frac{x}{(1-x^2)^\frac{1}{4}}$

$-\frac{2}{3}h(x)=\frac{x}{(1-x^2)^\frac{1}{4}}$

then using the quotient rule again:

$-\frac{2}{3}h'(x)=\frac{(1-x^2)^\frac{1}{4}+\frac{x^2}{2}\cdot (1-x^2)^{-\frac{3}{4}}}{(1-x^2)^\frac{1}{2}}$

multiplying by $(1-x^2)^\frac{3}{4}$ on top and bottom:

$-\frac{2}{3}h'(x)=\frac{(1-x^2)+\frac{x^2}{2}}{(1-x^2)^\frac{5}{4}}$

$-\frac{2}{3}h'(x)=\frac{(2-2x^2)+x^2}{2(1-x^2)^\frac{5}{4}}$

$h'(x)=-\frac{3(2-x^2)}{4(1-x^2)^\frac{5}{4}}$

$h'(x)=\frac{3}{4(1-x^2)^\frac{5}{4}}\cdot (x^2-2)$

tada

squance · « **Reply #3 on:** April 10, 2008, 11:10:01 pm »

Thanks...

But now I have more difficult questions....Please help me!!!

For the following functions evalutate the first few derivatives f'(x), f''(x), F^(3) (x) and so on. Deduce the pattern and write down an expression for the nth derivative f^(n)x

a). f(x) = x^n (answer is n! = n x (n-1) x (n-2) x (n-3)...

b). f(x) = 1/3x^3 (answer is (-1)^n (n+2)!/(6x^(n+3)

Assuming g(x) is twice differentiable, that is, we can write down g'(x) and g''(x), write down expressions for f''(x) if:

a). f(x) = xg(x^2) (answer is 6xg'(x^2) + 4x^3g''(x^2)

b). f(x) = g(root x) (answre is rootx g'' (root x) - g'(root x)/(4x root x)

Mao · « **Reply #4 on:** April 10, 2008, 11:17:26 pm »

part 1:

a)
$f(x)=x^n$

$\frac{df}{dx}=nx^{n-1}$

$\frac{d^2f}{dx^2}=(n)(n-1)x^{n-2}$

$\frac{d^3f}{dx^3}=(n)(n-1)(n-2)x^{n-3}$

so the overall pattern is (kth derivative):

$\frac{d^kf}{dx^k}=(n)(n-1)(n-2)\cdots(n-(k-1))x^{n-k}$

which simplifies to:

$\frac{d^kf}{dx^k}=\frac{n!}{(n-k)!}\cdot x^{n-k}$

so why is your answer like that... =S ?

*argh* do the rest 2morrow

Mao · « **Reply #5 on:** April 11, 2008, 03:59:48 pm »

And here's the rest:

part 1:
b)
$f(x)=\frac{1}{3x^3}=\frac{1}{3}x^{-3}$

$\frac{df}{dx}=-3 \cdot \frac{1}{3} \cdot x^{-4}$

$\frac{d^2f}{dx^2}=(-3)\cdot(-4)\cdot\frac{1}{3}\cdot x^{-5}=(-1)^2\cdot 3\cdot 4 \cdot \frac{1}{3}\cdot x^{-5}$

$\frac{d^3f}{dx^3}=(-3)\cdot(-4)\cdot (-5) \cdot \frac{1}{3} \cdot x^{-6} = (-1)^3 \cdot 3 \cdot 4 \cdot 5 \cdot \frac{1}{3} \cdot x^{-6}$

noticing the pattern here:
$\frac{d^nf}{dx^n}=(-1)^n \cdot \frac{(n+2)!}{2} \cdot \frac{1}{3} \cdot x^{-(3+n)}$

$\frac{d^nf}{dx^n}=\frac{(-1)^n \cdot (n+2)!}{6\cdot x^{3+n}}$

part 2:

Quote

Assuming g(x) is twice differentiable, that is, we can write down g'(x) and g''(x), write down expressions for f''(x) if:

a). f(x) = xg(x^2) (answer is 6xg'(x^2) + 4x^3g''(x^2)

b). f(x) = g(root x) (answre is rootx g'' (root x) - g'(root x)/(4x root x)

a)
using the product rule:

$\frac{df}{dx}=\left(\frac{d}{dx}x\right)\cdot g\left( x^2\right) + \left(\frac{d}{dx}g\left( x^2\right)\right) \cdot x$

$\frac{df}{dx}=g\left( x^2\right)+2x^2\cdot g'\left(x^2\right)$

therefore

$\frac{d^2f}{dx^2}=\frac{d}{dx}\left(g\left( x^2\right)\right)+\frac{d}{dx}\left(2x^2\cdot g'\left(x^2\right)\right)$

then using the product rule again:
$\frac{d^2f}{dx^2}=2x\cdot g'\left(x^2\right) + \left(\frac{d}{dx}2x^2\right)\cdot g'\left(x^2\right)\right) + \left(\frac{d}{dx}g'\left(x^2\right)\right)\cdot 2x^2$

$\frac{d^2f}{dx^2}=2x\cdot g'\left(x^2\right)+4x\cdot g'\left(x^2\right)+4x^3\cdot g''\left(x^2\right)$

$\frac{d^2f}{dx^2}=6x\cdot g'\left(x^2 \right) + 4x^3\cdot g''\left(x^2\right)$

b)

$f(x)=g\left(\sqrt{x}\right)$

$f(x)=g\left(x^{1/2}\right)$

using chain rule (let $u=x^{1/2}$ ):

$\frac{df}{dx}=\frac{d}{dx}\left(x^{1/2}\right)\cdot \frac{d}{du}g(u)$

$\frac{df}{dx}=\frac{1}{2}\cdot x^{-1/2} \cdot g' \left(\sqrt{x}\right)$

then, taking the $\frac{1}{2}$ out and using the product rule:

$\frac{d^2f}{dx^2}=\frac{1}{2}\cdot\frac{d}{dx}\left(x^{-1/2} \cdot g' \left(\sqrt{x}\right)\right)$

$\frac{d^2f}{dx^2}=\frac{1}{2}\cdot\left( \left(\frac{d}{dx}x^{-1/2}\right) \cdot g' \left(\sqrt{x}\right) + \left(\frac{d}{dx}g' \left(\sqrt{x}\right)\right) \cdot x^{-1/2}\right)$

$\frac{d^2f}{dx^2}=\frac{1}{2}\cdot \left( -\frac{1}{2}\cdot x^{-3/2}\cdot g' \left(\sqrt{x}\right) + \frac{1}{2}\cdot x^{-1/2} \cdot x^{-1/2} \cdot g''\left( \sqrt{x}\right)\right)$

$\frac{d^2f}{dx^2}=-\frac{1}{4}\cdot x^{-3/2}\cdot g' \left(\sqrt{x}\right) + \frac{1}{4}\cdot x^{-1} \cdot g''\left( \sqrt{x}\right)$

$\frac{d^2f}{dx^2}=\frac{g''\left( \sqrt{x}\right)}{4x} - \frac{g' \left(\sqrt{x}\right)}{4x\cdot \sqrt{x}}$

and why this is different to your solutions... i dont know
did i make a mistake somewhere?

squance · « **Reply #6 on:** April 11, 2008, 07:47:25 pm »

Quote from: Mao on April 11, 2008, 03:59:48 pm

b)

$f(x)=g\left(\sqrt{x}\right)$

$f(x)=g\left(x^{1/2}\right)$

using chain rule (let $u=x^{1/2}$ ):

$\frac{df}{dx}=\frac{d}{dx}\left(x^{1/2}\right)\cdot \frac{d}{du}g(u)$

$\frac{df}{dx}=\frac{1}{2}\cdot x^{-1/2} \cdot g' \left(\sqrt{x}\right)$

then, taking the $\frac{1}{2}$ out and using the product rule:

$\frac{d^2f}{dx^2}=\frac{1}{2}\cdot\frac{d}{dx}\left(x^{-1/2} \cdot g' \left(\sqrt{x}\right)\right)$

$\frac{d^2f}{dx^2}=\frac{1}{2}\cdot\left( \left(\frac{d}{dx}x^{-1/2}\right) \cdot g' \left(\sqrt{x}\right) + \left(\frac{d}{dx}g' \left(\sqrt{x}\right)\right) \cdot x^{-1/2}\right)$

$\frac{d^2f}{dx^2}=\frac{1}{2}\cdot \left( -\frac{1}{2}\cdot x^{-3/2}\cdot g' \left(\sqrt{x}\right) + \frac{1}{2}\cdot x^{-1/2} \cdot x^{-1/2} \cdot g''\left( \sqrt{x}\right)\right)$

$\frac{d^2f}{dx^2}=-\frac{1}{4}\cdot x^{-3/2}\cdot g' \left(\sqrt{x}\right) + \frac{1}{4}\cdot x^{-1} \cdot g''\left( \sqrt{x}\right)$

$\frac{d^2f}{dx^2}=\frac{g''\left( \sqrt{x}\right)}{4x} - \frac{g' \left(\sqrt{x}\right)}{4x\cdot \sqrt{x}}$

and why this is different to your solutions... i dont know
did i make a mistake somewhere?

My solutions are from the handbook so I don't know whether they are correct or not.

THanks again for the solutions.

I was wondering if you could give me a simple explanation of how implicit differentiation works...Im very confused...

AppleXY · « **Reply #7 on:** April 11, 2008, 07:55:59 pm »

Implicit differentiation is where you diff. all sides of the equation

and subject dy/dx.

squance · « **Reply #8 on:** April 11, 2008, 07:58:32 pm »

Quote from: AppleXY on April 11, 2008, 07:55:59 pm

Implicit differentiation is where you diff. all sides of the equation and subject dy/dx.

Do u think you could give me an example of how this works???

Mao · « **Reply #9 on:** April 11, 2008, 08:51:52 pm »

implicit differentiation is basically differentiating against y then add a dy/dx to the end:

$\frac{d}{dx}y^2=\left(\frac{d}{dy}y^2\right) \cdot \frac{dy}{dx}=2y\cdot \frac{dy}{dx}$

with that, you can rearrange to find dy/dx.

such as:

$y^2=x^3-1$

$\frac{d}{dx}y^2=\frac{d}{dx}\left(x^3-1\right)$

$\frac{d}{dy}y^2 \cdot \frac{dy}{dx}=3x^2$

$2y\cdot \frac{dy}{dx}=3x^2$

$\frac{dy}{dx}=\frac{3x^2}{2y}$

AppleXY · « **Reply #10 on:** April 11, 2008, 09:46:45 pm »

Yeah, you can use the chain rule to find $\frac{d}{dx} y^{2} = 2y\cdot \frac{dy}{dx}$ for example, i.e. to implicitly differentiate.

However, remember to use the most efficient way of differentiating.

(eg. don't implicitly diff x + y = 1 as y = 1 - x :. $\frac{dy}{dx}$ = -1)

squance · « **Reply #11 on:** April 12, 2008, 11:48:05 pm »

Thanks. I was able to do some of the simple implicit questions.

But then....I got stuck on these..can anyone figure this out for me....

loge(x/y) = cot(xy) (answer is y +y^x cosec^2 xy/x-x^2cosec^2 xy)

cos(x-y) = xe^x (for this one I got e^x(1+x)/(-sin(x-y)) but the answer says its 1 + e^x(1+x)/(sin(x-y))

Mao · « **Reply #12 on:** April 13, 2008, 08:52:50 am »

$\frac{d}{dx}ln\left(\frac{x}{y}\right)=\frac{d}{dx}\cot (x\cdot y)$

using the chain rule on the left hand side:

$\frac{d}{dx}ln \left( \frac{x}{y}\right)$

$\implies \frac{d}{dx}\left( \frac{x}{y}\right) \cdot \left(\frac{x}{y}\right)^{-1}$

using the quotient rule and implicitly differentiating y:

$\implies \frac{y-x\cdot \frac{dy}{dx}}{y^2} \cdot \frac{y}{x}$

$\implies \frac{y-x \cdot \frac{dy}{dx}}{x\cdot y}$

for the RHS:
the derivative of $\cot (x)$ is $-\frac{1}{\sin^2 x}$

using the chain rule:

$\frac{d}{dx}\cot(x\cdot y)$

$\implies \frac{d}{dx}(x\cdot y) \cdot \left(-\frac{1}{\sin^2 (x \cdot y )}\right)$

then using the product rule:
$\implies \left(y+x\cdot \frac{dy}{dx}\right) \cdot \left(-\frac{1}{\sin^2 (x \cdot y )}\right)$

$\implies -\frac{y+x\cdot \frac{dy}{dx}}{\sin^2 (x \cdot y )}$

equating the LHS and RHS:

$\frac{y-x \cdot \frac{dy}{dx}}{x\cdot y}=-\frac{y+x\cdot \frac{dy}{dx}}{\sin^2 (x \cdot y )}$

cross multiplying:

$\left( y-x\cdot \frac{dy}{dx}\right) \cdot \sin^2 (x \cdot y ) = - \left( y+x\cdot \frac{dy}{dx} \right) \cdot x \cdot y$

$y \cdot \sin^2 (x \cdot y ) - x \cdot \sin^2 (x \cdot y ) \cdot \frac{dy}{dx} = -x\cdot y^2 - x^2 \cdot y \cdot \frac{dy}{dx}$

$y \cdot \sin^2 (x \cdot y ) + x\cdot y^2 = x \cdot \sin^2 (x \cdot y ) \cdot \frac{dy}{dx} - x^2 \cdot y \cdot \frac{dy}{dx}$

$y \cdot \sin^2 (x \cdot y ) + x\cdot y^2 = \left(x \cdot \sin^2 (x \cdot y )-x^2 \cdot y\right) \cdot \frac{dy}{dx}$

$\frac{dy}{dx}=\frac{y \cdot \sin^2 (x \cdot y ) + x\cdot y^2}{x \cdot \sin^2 (x \cdot y )-x^2 \cdot y}$

which is not algebraicly equivalent to the solution you gave...

did i make a mistake up there somewhere?

*and now i need to catch a train*

dcc · « **Reply #13 on:** April 13, 2008, 06:39:02 pm »

$cos(x - y) = xe^x$

Consider:

$u = x - y$

$\dfrac{du}{dx} = 1 - \dfrac{dy}{dx}$

Therefore, using the chain rule (for the left hand side) to implicitly differentiate (i.e $\dfrac{d}{du}(\cos u) \times \dfrac{du}{dx}$ )

$(1 - \dfrac{dy}{dx})(-sin(x - y)) = e^x + xe^x$

$(\dfrac{dy}{dx} - 1)(\sin (x - y)) = e^x(1 + x)$

$\dfrac{dy}{dx} - 1 = \dfrac{e^x(1 + x)}{\sin (x - y)}$

$\dfrac{dy}{dx} = 1 + \dfrac{e^x(1 + x)}{\sin (x - y)}$

dcc · « **Reply #14 on:** April 13, 2008, 07:23:03 pm »

A tip for logarithms in general:

If a question asked you to do something like

$\mbox{If $y = \log_e\dfrac{2x^2 + 4x + 3}{3x + 9}$, then find $\dfrac{dy}{dx}$}$

Instead of having to use the quotient + chain rules, it is easier to just simply write:

$y = log_e(2x^2 + 4x + 3) - log_e(3x + 9)$

and then differentiate each term separately (thus eliminating the annoyingness of the quotient rule)

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squance

Differentiation...

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