Understanding Derivatives and Integrals: A Basic Guide for Beginners

[Mol_Bolom]

I just want to see if I got this correct. From what all I've read it seems that I have most of it understood, but eh, I don't trust my judgement...

Lets say we have f(x) = {{3x^3 + 8x^2 + 7x + 12} \over {4x^2 - 12x - 15}}

And the derivative...

<br /> {d \over dx} f(x) = \lim _{h \rightarrow 0} {{f(x+h) - f(x)} \over h} =<br /> {{<br /> {d \over dx} (3x^3 + 8x^2 + 7x + 12)<br /> }<br /> \over<br /> {<br /> {d \over dx} (4x^2 - 12x - 15)<br /> }} = f&#039;(x)<br />

Thus the integral would be...

<br /> \int {f&#039;(x)} \textbf{ }dx = f(x)<br />

And if the constants are unknown, thus letting a and b represent the constants...

<br /> \int {f&#039;(x)} \text{ } dx = {{3x^3 + 8x^2 + 7x + a} \over {4x^2 - 12x + b}}<br />

 

[Hootenanny]

I just want to see if I got this correct. From what all I've read it seems that I have most of it understood, but eh, I don't trust my judgement...

Lets say we have f(x) = {{3x^3 + 8x^2 + 7x + 12} \over {4x^2 - 12x - 15}}

And the derivative...

<br /> {d \over dx} f(x) = \lim _{h \rightarrow 0} {{f(x+h) - f(x)} \over h} =<br /> {{<br /> {d \over dx} (3x^3 + 8x^2 + 7x + 12)<br /> }<br /> \over<br /> {<br /> {d \over dx} (4x^2 - 12x - 15)<br /> }} = f&#039;(x)<br />

Thus the integral would be...

<br /> \int {f&#039;(x)} \textbf{ }dx = f(x)<br />

And if the constants are unknown, thus letting a and b represent the constants...

<br /> \int {f&#039;(x)} \text{ } dx = {{3x^3 + 8x^2 + 7x + a} \over {4x^2 - 12x + b}}<br />

I'm afraid that what you have written isn't correct. You might want to have a read of these tutorials for https://www.physicsforums.com/showthread.php?t=139690".

 

[HallsofIvy]

I just want to see if I got this correct. From what all I've read it seems that I have most of it understood, but eh, I don't trust my judgement...

Lets say we have f(x) = {{3x^3 + 8x^2 + 7x + 12} \over {4x^2 - 12x - 15}}

And the derivative...

<br /> {d \over dx} f(x) = \lim _{h \rightarrow 0} {{f(x+h) - f(x)} \over h} =<br /> {{<br /> {d \over dx} (3x^3 + 8x^2 + 7x + 12)<br /> }<br /> \over<br /> {<br /> {d \over dx} (4x^2 - 12x - 15)<br /> }} = f&#039;(x)<br />

No. the derivative of f(x)/g(x) is (f'(x)g(x)- f(x)g'(x))/g²(x), not f'/g'.

Thus the integral would be...

<br /> \int {f&#039;(x)} \textbf{ }dx = f(x)<br />

And if the constants are unknown, thus letting a and b represent the constants...

<br /> \int {f&#039;(x)} \text{ } dx = {{3x^3 + 8x^2 + 7x + a} \over {4x^2 - 12x + b}}<br />

Nor can you integrate f/g by integrating numerator and denominator separately.

 

[rzaidan]

Hi Mol_Bolom
when you find the derivative by definition
(d/dx)(f(x)=lim(f(x+h)-f(x))/h you can find the limit of d/dh(f(x+h)-f(x))
h__ 0 ---------------
d/dh(h)
for example if F(x)=x^2 you have

d/dx(F(x))=lim ((x+h)^2-x^2)/h =
h___0
=lim [2(x+h)-0]/1 by diff. both num. and denum. w.r.t h and take the limit as h
h___0
approaches 0 you get
d/dx(F(x))=2x
Best wishes
Riad Zaidan