Differentiate y = 5x^2 - (2x+1)^3

[Monkeyseat]

Question:

Differentiate y = 5x^2 - (2x+1)^3

------------

Here is how I have been doing these. For example: y = 4e^(3-x).

Here I made u = 3 - x and y = 4e^u.

du/dx = -1 and dy/du = 4e^u.

dy/dx = 4e^u * -1
dy/dx = -4e^(3-x)

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Okay, so back on to y = 5x^2 - (2x+1)^3.

What do I choose for u and y? I chose u = 2x + 1 and y = 5x^2 - u^3 but it went wrong.

So du/dx = 2 and dy/du = 10x - 3u^2.

dy/dx = 2(10 - 3u^2)
dy/dx = 20 - 6u^2
dy/dx = 20 - 6(2x+1)^2

I know this is wrong, the answer is meant to be 10 - 6(2x+1)^2 but I can't get this! I think I am choosing the wrong values for u and v but I don't know what else to use. I think its because you can't have y = 10x - 3u^2 as it has three variables? I only know how to do it this way at the moment.

Any help is appreciated.

 

[Deleted member 4993]

Question:

Differentiate y = 5x^2 - (2x+1)^3

Here is how I have been doing these. For example: y = 4e^(3-x)

Here I made u = 3 - x and y = 4e^u.

du/dx = -1 and dy/du = 4e^u.

dy/dx = 4e^u * -1
dy/dx = -4e^(3-x)

Okay, so back on to y = 5x^2 - (2x+1)^3.

What do I choose for u and y? I chose u = 2x + 1 and y = 5x^2 - u^3 but it went wrong.

In this problem you don't need 'u' substitution.

Just use "power rules"

\(\displaystyle \frac{dx^n}{dx} \, = \, n\cdot x^{n-1}\)

and the "chain rule" (you could call this substitution - but that's stretching it)

\(\displaystyle \frac{d f[g(x)]}{dx} \, = \, \frac{d[f(g)]}{dg} \cdot \frac{d[g(x)]}{dx}\)

Do the differentiation "part by part"

\(\displaystyle \frac{d(5x^2)}{dx} \, = \, 10 \cdot x\)

and

for the next part

g(x) = 2x+1

\(\displaystyle \frac{dg}{dx} \, = \, 2\)

f(g) = g^3

\(\displaystyle \frac{df}{dg} \, = \, 3\cdot g^2 \, = \, 3\cdot (2x \, + \, 1)^2\)

putting all these together

\(\displaystyle \frac{d[5x^2 \, - \, (2x+1)^3]}{dx} \, = \, 10 \cdot x \, - \, 6\cdot (2x \, + \, 1)^2\)...corrected

That's all.....




So du/dx = 2 and dy/du = 10x - 3u^2.

dy/dx = 2(10 - 3u^2)
dy/dx = 20 - 6u^2
dy/dx = 20 - 6(2x+1)^2

I know this is wrong, the answer is meant to be 10 - 6(2x+1)^2 but I can't get this! I think I am choosing the wrong values for u and v but I don't know what else to use. I think its because you can't have y = 10x - 3u^2 as it has three variables? I only know how to do it this way at the moment.

Any help is appreciated.

 

[Monkeyseat]

Question:

Differentiate y = 5x^2 - (2x+1)^3

Here is how I have been doing these. For example: y = 4e^(3-x)

Here I made u = 3 - x and y = 4e^u.

du/dx = -1 and dy/du = 4e^u.

dy/dx = 4e^u * -1
dy/dx = -4e^(3-x)

Okay, so back on to y = 5x^2 - (2x+1)^3.

What do I choose for u and y? I chose u = 2x + 1 and y = 5x^2 - u^3 but it went wrong.

In this problem you don't need 'u' substitution.

Just use "power rules"

\(\displaystyle \frac{dx^n}{dx} \, = \, n\cdot x^{n-1}\)

and the "chain rule" (you could call this substitution - but that's stretching it)

\(\displaystyle \frac{d f[g(x)]}{dx} \, = \, \frac{d[f(g)]}{dg} \cdot \frac{d[g(x)]}{dx}\)

Do the differentiation "part by part"

\(\displaystyle \frac{d(5x^2)}{dx} \, = \, 10 \cdot x\)

and

for the next part

g(x) = 2x+1

\(\displaystyle \frac{dg}{dx} \, = \, 2\)

f(g) = g^3

\(\displaystyle \frac{df}{dg} \, = \, 3\cdot g^2 \, = \, 3\cdot (2x \, + \, 1)^2\)

putting all these together

\(\displaystyle \frac{d[5x^2 \, - \, (2x+1)^3]}{dx} \, = \, 10 \cdot x \, + \, 6\cdot (2x \, + \, 1)^2\)

That's all.....




So du/dx = 2 and dy/du = 10x - 3u^2.

dy/dx = 2(10 - 3u^2)
dy/dx = 20 - 6u^2
dy/dx = 20 - 6(2x+1)^2

I know this is wrong, the answer is meant to be 10 - 6(2x+1)^2 but I can't get this! I think I am choosing the wrong values for u and v but I don't know what else to use. I think its because you can't have y = 10x - 3u^2 as it has three variables? I only know how to do it this way at the moment.

Any help is appreciated.

That's kind of what I thought just after posting the original message.

I did:

u = 2x + 1
du/dx = 2

y = u^3
dy/du = 3u^2

So:

dy/dx = (du/dx * dy/du) + 10x
dy/dx = 6u^2 +10x
dy/dx = 6(2x+1)^2 + 10x

Is that ok? So I was along the right lines by differentiating 5x^2 separately.

 

[Deleted member 4993]

[
Is that ok? So I was along the right lines by differentiating 5x^2 separately.

Yes

f(x) = u(x) ± v(x)

then

f'(x) = u'(x) ± v'(x)

 

[Monkeyseat]

[
Is that ok? So I was along the right lines by differentiating 5x^2 separately.

Yes

f(x) = u(x) ± v(x)

then

f'(x) = u'(x) ± v'(x)

Thanks for helping. Just one last thing.

So for y = 5x^2 - (2x + 1)^3,

u = 2x + 1
du/dx = 2

u = -u^3
du/dx = -3u^2 <--- Is this not negative?

Would it not be negative like I did above because it is "- (2x + 1)^3"? On yours it is positive.

That would give a different answer.

dy/dx = 10x - 6u^2
dy/dx = 10x - 6(2x+1)^2

 

[Deleted member 4993]

[
Is that ok? So I was along the right lines by differentiating 5x^2 separately.

Yes

f(x) = u(x) ± v(x)

then

f'(x) = u'(x) ± v'(x)

Thanks for helping. Just one last thing.

So for y = 5x^2 - (2x + 1)^3,

u = 2x + 1
du/dx = 2

u = -u^3
du/dx = -3u^2 <--- Is this not negative?

Would it not be negative like I did above because it is "- (2x + 1)^3"? On yours it is positive.

That would give a different answer.

dy/dx = 10x - 6u^2
dy/dx = 10x - 6(2x+1)^2 ...........Correct ....good catch ... I fixed my original response

 

[Monkeyseat]

Thanks again for helping.