Differential calculus for univariate models

Titicamara · 2006-08-14 17:56:22

Please help me with the following questions. Thank you very much.

A. Find the derivative for each of the following function.

1. y=x^4 + 2e^x
2. y=10lnx
3. y= (1.08)^x + 10

B. Use the product rule or the quotient rule to find the derivative of each of the following functions.

1. y=5 + xe^x
2. y=x^2e^x + xlnx
3. y=5e^x lnx
4. y=x^5/6 e^x

C. Find the derivative of each of the following functions using chain or function of a function' rule.

1. y= (4+lnx^2)^-1
2. y=10x + e^lnx
3. y=10 / (1-5e^0.2x)
4. y=100 - 5e^-0.3x

And please show me the workings too so that I can revise better. Thank you.

mathsyperson · 2006-08-14 23:28:37

A.

1. Just differentiate the 2 terms separately and add them afterwards.
y' = 4x³ + 2e[sup]x[/sup]

2. d(ln x)/dx = 1/x, so y' = 10/x

3. d(a[sup]x[/sup])/dx = a[sup]x[/sup] * ln a, so y' = (1.08)[sup]x[/sup] * ln 1.08.

B.

For the product rule, you just write the first term and differentiate the second, and then do the opposite.

1. d(x)/dx = 1 and d(e[sup]x[/sup])/dx = e[sup]x[/sup], so d(xe[sup]x[/sup])/dx = e[sup]x[/sup] + xe[sup]x[/sup]. The 5 differentiates to nothing.

2. I'm presuming the first term there is x²e[sup]x[/sup], rather than x[sup]2e^x[/sup], because that would be too complicated for just the product rule.

If so, the product rule can show that d(x²e[sup]x[/sup])/dx = 2xe[sup]x[/sup] + x²e[sup]x[/sup] and that d(xlnx) = 1 + ln x. Therefore, y' = xe[sup]x[/sup](2+x) + 1 + ln x.

3. d(5e[sup]x[/sup])/dx = 5e[sup]x[/sup] and d(ln x)/dx = 1/x, so y' = 5e[sup]x[/sup](ln x + 1/x)

4. Using the same method as in the others, y' = e[sup]x[/sup](x[sup]5/6[/sup] + 5/6x[sup]-1/6[/sup])

C.

1. The chain rule works by splitting a complicated expression into a number of simpler ones and differentiating them separately. In this case, (4+lnx²) becomes t.

So, dy/dt = -t^-2 and dt/dx = 2/x. Therefore, dy/dx = -2/(t²x) = -2/(x(4+ln x²)²).

2. The 10x is trivial, so the chain rule is only needed for e^ln x. However, it can be seen without the chain rule that the ln and the e cancel each other out, leaving just x. Therefore, dy/dx = 11.

3. Taking t as (1-5e^0.2x), we can see that dy/dt = -10/t² and dt/dx = -e^0.2x.
Therefore, dy/dx = (10e^0.2x)/t² = (10e^0.2x)/(1-5e^0.2x)².

4. The 100 is trivial, so we are only interested in the -5e^-0.3x.
The chain rule is not needed here, because we can use the identity that d(e^ax)/dx = ae^ax.

Therefore, y' = 1.5e^0.3x.

Titicamara · 2006-08-15 03:45:35

All your answers are correct but may I ask a few questions. I am very weak in maths.

A.

1. y' = 4x³ + 2e^x

I understand that 4x^3 is from 4 * x ^4-1 but what about 2e^x? How did you get 2e^x? In the future whenever I see e what do I do with it?

B.

1. How do you get d(x)/dx = 1 and d(ex)/dx = ex? I understood how 5 differentiates to 0 because 5 = 5x^0.

I am vague especially when it comes to e and ln. How am I to derive them?

C.

2. How is it that ln and the e cancel each other out?

I totally do not get question 3 and 4.

Here is how I did my question 3 and 4 which I am stucked at the moment.

3. y=10 / (1-5e^0.2x)
y=10 * (1-5e^0.2x)^-1

u = 10 * (1-5e^0.2x)
y = u^-1

du/dx = -5e^0.2x dy/dx = -1u^-2

dy/dx = dy/du * du/dx

= ( -1u^-2) (-5e^0.2x)
= -1 (10) (1-5e^0.2x)^-2 * ( -5e^0.2x)

After this step, I am lost as to how to continue on.

4.

y = 100-5e^-0.3x

u= 100-5e^-0.3x
y= u^-0.3

du/dx = -5e^-0.3x dy/dx = -0.3u^-1.3

dx/dy = dy/du * du/dx
=-0.3u^-1.3 * -5e^-0.3x
= -0.3 ( 100 - 5e^-0.3x)^-1.3 * -5e^-0.3x

And if I to continue on my answer will be completely different from yours.

Please advise. Thank you

Last edited by Titicamara (2006-08-15 03:47:13)

mathsyperson · 2006-08-15 08:39:08

A1. e^x is one of the easiest things to differentiate because the differential of e^x is e^x.
Multiplying by a constant doesn't make any difference, and so d(2e^x)/dx = 2e^x.

B1. From above, d(e^x)/dx = e^x. You can work out how x differentiates to 1 by thinking about it. x is the same as x^1, and so you multiply by the power and then take 1 off it, to give 1x^0, which is equivalent to 1.

C2. ln and e cancel each other out because they are opposites. If you take a number, raise it to the power of e, and find the ln of your answer, then you'll get your original number back.

C3. You've made a slight mistake here, because by saying that y = u^-1, you're saying that y is 1/10(1-5e^0.2x), which means that your answer is 100 times too small.

Also, your value of du/dx should be -10e^0.2x.
Try it again, taking u as (1-5e^0.2x) and y as 10/u.

C4. I'm not sure what you've done here, because your expression for u is the same as your expression for y, which means that you haven't actually used the chain rule at all.

You can ignore the 100 because it will differentiate to nothing, so just focus on the -5e^-0.3x.
If you turn the -0.3x into u, then you just need to differentiate -5e^u and -0.3x, which are both fairly simple.

Math Is Fun Forum

#1 2006-08-14 17:56:22

Differential calculus for univariate models

#2 2006-08-14 23:28:37

Re: Differential calculus for univariate models

#3 2006-08-15 03:45:35

Re: Differential calculus for univariate models

#4 2006-08-15 08:39:08

Re: Differential calculus for univariate models

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