MHB Differentiation Techniques in Calculus

[ardentmed]

Hey guys,

I have a couple of questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For 1a, I just separated the function into single terms with √x as the denominator. This ultimately resulted in: (5/2)x^(3/2) - (9/2)√x + 5/(2x^[3/2])

For 1b, I multiplied the outer variable into the expression in the bracket. This gave me: (4/3) w^(1/3) + (2/(3w^[5/3])) + 2e^w * (w^1/3 + 1/(3w^(2/3))

Does that look right to you guys?

As for 2a, I used the product rule and got 4x^3 *e^(-1/(x^2)) +3xe^(-1/(x^2)) - 2/x^2 * ln (2/x) + 2/(x^2)

Finally, for 2b, I just took the derivative as per usual and obtained:

(-3sin3Ø/cos3Ø) - (sin(ln(3Ø))/Ø)Thanks in advance.

 

[anemone]

For 1a, I just separated the function into single terms with √x as the denominator. This ultimately resulted in: (5/2)x^(3/2) - (9/2)√x + 5/(2x^[3/2])

Correct.

For 1b, I multiplied the outer variable into the expression in the bracket. This gave me: (4/3) w^(1/3) + (2/(3w^[5/3])) + 2e^w * (w^1/3 + 1/(3w^(2/3))

Correct.

As for 2a, I used the product rule and got 4x^3 *e^(-1/(x^2)) +3xe^(-1/(x^2)) - 2/x^2 * ln (2/x) + 2/(x^2)

I got $4x^3e^{-(1/x^2)}+2xe^{-(1/x^2)}-\dfrac{2}{x^2}-\dfrac{2}{x^2}\ln \dfrac{2}{x}$.

You need to show us your work before we can tell you where you went wrong...

 

[ardentmed]

Correct.
Correct.

I got $4x^3e^{-(1/x^2)}+2xe^{-(1/x^2)}-\dfrac{2}{x^2}-\dfrac{2}{x^2}\ln \dfrac{2}{x}$.

You need to show us your work before we can tell you where you went wrong...

Alright, I re-did the question and got the same answer.

2b is correct, right? I just applied the product rule.

Thanks again.

 

[SuperSonic4]

Alright, I re-did the question and got the same answer.

2b is correct, right? I just applied the product rule.

Thanks again.

For 2b you need to apply the chain rule twice. The Product rule is when you have two expressions multiplied (ie $$\dfrac{d}{dx} \sin(x)\cos(x)$$

Since you are adding you can treat each expression by itself and add the result

$$s(\theta) = \ln(\cos(3\theta)) + \cos(\ln(3\theta))$$

$$s'(\theta) = \dfrac{d}{d\theta}\left(\ln(\cos(3\theta))\right) + \dfrac{d}{d\theta}\left(\cos(\ln(3\theta))\right)$$

 

[ardentmed]

For 2b you need to apply the chain rule twice. The Product rule is when you have two expressions multiplied (ie $$\dfrac{d}{dx} \sin(x)\cos(x)$$

Since you are adding you can treat each expression by itself and add the result

$$s(\theta) = \ln(\cos(3\theta)) + \cos(\ln(3\theta))$$

$$s'(\theta) = \dfrac{d}{d\theta}\left(\ln(\cos(3\theta))\right) + \dfrac{d}{d\theta}\left(\cos(\ln(3\theta))\right)$$

So the final answer just: (-3sin3Ø/cos3Ø) - (sin(ln(3Ø))/Ø) After applying the chain rule?

 

[anemone]

So the final answer just: (-3sin3Ø/cos3Ø) - (sin(ln(3Ø))/Ø) After applying the chain rule?

Yeap!(Yes)