Would the first derivative of 3x(5x^2 + 1) be: 15x^2 +1?

[Umbra]

Would the first derivative of 3x(5x^2 + 1) be:
15x^2 +1?

3x(5x^2) = 15x^3 -^1 = 15x^2 +1

 

[stapel]

Would the first derivative of 3x(5x^2 + 1) be:
15x^2 +1?

No.

3x(5x^2) = 15x^3 -^1 = 15x^2 +1

I'm sorry, but I don't understand what you're doing here...? For instance, what is the meaning of "-^1"? How did you go from one term to two?

In other words, you'd started with this:

. . . . .\(\displaystyle 3x\, (5x^2\, +\, 1)\)

Presumably, the instructions told you to "find the derivative" of the expression. Either you used the Product Rule, or else you multiplied through and used the Power Rule for polynomials. The latter method would be easier, of course, so I'll guess that you used that method:

. . . . .\(\displaystyle 3x\, (5x^2\, +\, 1)\)

. . . . .\(\displaystyle 3x\, (5x^2)\, +\, 3x\, (1)\)

. . . . .\(\displaystyle 15x^3\, +\, 3x\)

Then you differentiated. But how did you arrive at your value?

Please show all of your steps. Thank you!

 

[Umbra]

No.


I'm sorry, but I don't understand what you're doing here...? For instance, what is the meaning of "-^1"? How did you go from one term to two?

In other words, you'd started with this:

. . . . .\(\displaystyle 3x\, (5x^2\, +\, 1)\)

Presumably, the instructions told you to "find the derivative" of the expression. Either you used the Product Rule, or else you multiplied through and used the Power Rule for polynomials. The latter method would be easier, of course, so I'll guess that you used that method:

. . . . .\(\displaystyle 3x\, (5x^2\, +\, 1)\)

. . . . .\(\displaystyle 3x\, (5x^2)\, +\, 3x\, (1)\)

. . . . .\(\displaystyle 15x^3\, +\, 3x\)

Then you differentiated. But how did you arrive at your value?

Please show all of your steps. Thank you!

So here's the question: Determine the equations of the tangent lines to the graph of f(x) = 3x(5x^2 + 1) that are parallel to the line y = 8x + 9. Sketch a graph of the function, the tangent lines, and the line y = 8x + 9.

and my steps:
f(x) = 3x(5x^2 + 1)
= 15x^3 + 3x
f'(x) = 45x^2 + 3

Then matching the slope to the parallel line:
8 = 45x^2 + 3
8 - 3 = 45x^2
5 = 45x^2
45/5 = x^2
9 = x^2
+ or - sqrt(9) = x
x = 3 or x = -3

Would that all be correct?

I just feel like what I've done is wrong based on these results.

Thanks for the response! Kind regards.

 

[stapel]

I figured 3x x 5x^2 would be 15x^3 + 3x

No; the product of these terms is not a sum. To learn now to multiply polynomials, try here.

then the derivative of such would become 15x^2 +3x?

No; The derivative of 3x is not 3x. To learn how to differentiate polynomial terms (among other formulas), try here.

So here's the question: Determine the equations of the tangent lines to the graph of f(x) = 3x(5x^2 + 1) that are parallel to the line y = 8x + 9. Sketch a graph of the function, the tangent lines, and the line y = 8x + 9.

and my steps:
f(x) = 3x(5x^2 + 1)
= 15x^3 + 3x
f'(x) = 45x^2 + 3

Correct.

Then matching the slope to the parallel line:
8 = 45x^2 + 3
8 - 3 = 45x^2
5 = 45x^2

I just feel like what I've done is wrong based on these results.

What aspect of this above result leads you to "feel like what you've done is wrong"?

I just did:
5 = 45x^2
45/5 = x^2
9 = x^2
+ or - sqrt(9) = x
x = 3 or x = -3

Would that all be correct?

What value do you get for m = f'(x) at these values of x? If you obtain m = 8, then you have the correct points. If not, then you don't.

By the way, how did you go from here:

. . . . .\(\displaystyle 5\, =\, 45\, x^2\)

. . . . .\(\displaystyle \dfrac{5}{45}\, =\, \dfrac{45\, x^2}{45}\)

...to:

. . . . .\(\displaystyle 9\, =\, x^2\)

:shock:

 

[Umbra]

No; the product of these terms is not a sum. To learn now to multiply polynomials, try here.


No; The derivative of 3x is not 3x. To learn how to differentiate polynomial terms (among other formulas), try here.


Correct.


What aspect of this above result leads you to "feel like what you've done is wrong"?


What value do you get for m = f'(x) at these values of x? If you obtain m = 8, then you have the correct points. If not, then you don't.

By the way, how did you go from here:

. . . . .\(\displaystyle 5\, =\, 45\, x^2\)

. . . . .\(\displaystyle \dfrac{5}{45}\, =\, \dfrac{45\, x^2}{45}\)

...to:

. . . . .\(\displaystyle 9\, =\, x^2\)

:shock:

Okay... I'm dumb... I did 45/5 = 9... Not 5/45 = 0.111
0.111 = x^2
+ or - sqrt (0.111) = x
x = 0.333 or x = -0.333

f'(0.333) = 45(0.333)^2 + 3 = 8...

God...

Thank you so much!
Kind regards.

 

[stapel]

Okay... I'm dumb... I did 45/5 = 9... Not 5/45 = 0.111

Unless you're told to use decimal approximations, don't use them! In general, the calculator digits will not be counted as a "correct" answer.

The value of 5/45 is 1/9. The square root of 1/9 is 1/3 (plus or minus, when you're solving, as here). Use these values! :shock: