I have to differentiate this

[cotfw]

X/((1+x²)^.5)

i tried several times but my answer was never the same as the book.

 

[Deleted member 4993]

X/((1+x²)^.5)

i tried several times but my answer was never the same as the book.

Note that your numerator is "X" (capital "X")

and the denominator is "x" - two different animals.

Since you have tried several times - share at least one of those - and we will help you find your mistake.

Please share your work with us ...

If you are stuck at the beginning tell us and we'll start with the definitions.

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[cotfw]

Ok I'll post it. I only didn't post my work because I'm using an iPad and I'm not used to typing with it. The uppercase x was just he ipad capitalizing for me.


Since the equation is the same as x/1 * 1/(1+x²)^1/2, I thought I would differentiate both the left and right, then find the differential of the product.

For the right, I came to -x(1+x²)^-3/2

For or the left, I came to 1.

for the product of he differentials, I came to (-x²(1+x)^1/2 + (1+x²)^3/2)/ (1+x)²

Which is incorrect. Do you know how to solve this?

 

[Ishuda]

When you have one function times another, you use the product rule for differentiation which says
(fg)' = f' g + f g'
where the ' indicates differentiation. You have the proper derivatives [edit, see below] but have some mistakes/typos. So
f = x
g = 1/(1+x²)^1/2 = (1+x²)^-1/2
f' = 1
g' = \(\displaystyle \frac{-2 x}{2 (1+x^2)^{\frac{3}{2}}}\) = -\(\displaystyle x\space (1+x^2)^{-\frac{3}{2}}\)

Now put them together with the product differentiation rule. If you need to simplify the answer, do that also.
f' g + f g' = 1 * (1+x²)^-1/2 + x (-\(\displaystyle x\space (1+x^2)^{-\frac{3}{2}}\))
= (1+x²)^-1/2 - x² (1+x²)^-3/2 = (1+x²)^-3/2 [ (1 + x²) - x² ] = (1+x²)^-3/2


You can also get the same answer with the division rule, i.e.

(f/g)' = (f' g - f g') / g²
In that case f would be the same as above but g would change
f = x
g = (1+x²)^1/2

Edit: Actually, looking at this again, it looks like your answer may be correct (if you fix the minor mistakes/typos) but needs simplification. Put back the x² instead of x in the proper places and factor out (1+x²)^1/2 or multiply numerator and denominator by (1+x²)^1/2.

 

[cotfw]

When you have one function times another, you use the product rule for differentiation which says
(fg)' = f' g + f g'
where the ' indicates differentiation. You have the proper derivatives [edit, see below] but have some mistakes/typos. So
f = x
g = 1/(1+x²)^1/2 = (1+x²)^-1/2
f' = 1
g' = \(\displaystyle \frac{-2 x}{2 (1+x^2)^{\frac{3}{2}}}\) = -\(\displaystyle x\space (1+x^2)^{-\frac{3}{2}}\)

Now put them together with the product differentiation rule. If you need to simplify the answer, do that also.
f' g + f g' = 1 * (1+x²)^-1/2 + x (-\(\displaystyle x\space (1+x^2)^{-\frac{3}{2}}\))
= (1+x²)^-1/2 - x² (1+x²)^-3/2 = (1+x²)^-3/2 [ (1 + x²) - x² ] = (1+x²)^-3/2


You can also get the same answer with the division rule, i.e.

(f/g)' = (f' g - f g') / g²
In that case f would be the same as above but g would change
f = x
g = (1+x²)^1/2

Edit: Actually, looking at this again, it looks like your answer may be correct (if you fix the minor mistakes/typos) but needs simplification. Put back the x² instead of x in the proper places and factor out (1+x²)^1/2 or multiply numerator and denominator by (1+x²)^1/2.

thanks that was really clear and helpful : )

 

[Deleted member 4993]

Ok I'll post it. I only didn't post my work because I'm using an iPad and I'm not used to typing with it. The uppercase x was just he ipad capitalizing for me.


Since the equation is the same as x/1 * 1/(1+x²)^1/2, I thought I would differentiate both the left and right, then find the differential of the product.

For the right, I came to -x(1+x²)^-3/2

For or the left, I came to 1.

for the product of he differentials, I came to (-x²(1+x²)^1/2 + (1+x²)^3/2)/ (1+x)²

Which is incorrect. Do you know how to solve this?

One more step (may be) is needed:

(-x²(1+x²)^1/2 + (1+x²)^3/2)/ (1+x)² = (1+x²)^1/2[(-x² + (1+x²))]/ (1+x)² = (1+x²)^1/2/(1+x)²