Differentiate f(x)= (8-2x)^3(x^2+1)^5.
My ans is 2(8-2x)^2(x^2+1)^4(-3x^2-10x+37). Plz recheck my simplify ans.
-my simplify ans is not the same as wolfram but the long ans is same with wolfram-
f' = 3(8-2x)^2 (-2) (x^2+1)^5 + 5(8-2x)^3 (x^2+1)^4 (2x)
= (8-2x)^2 (x^2+1)^4 (-6(x^2+1) + 5(8-2x)(2x))
= (8-2x)^2 (x^2+1)^4 (26x^2 - 80x + 6)
= -8(4-x)^2 (x^2+1)^4 (13x^2 - 40x + 3)
which is what Wolfram gets.
Watch for those pesky chain rule factors and minus signs.
I still don't get it how do u simplify from
= (8-2x)^2 (x^2+1)^4 (26x^2 - 80x + 6) to
= -8(4-x)^2 (x^2+1)^4 (13x^2 - 40x + 3) ... ?
What can i conclude from it is (4-x)^2 &(13x^2 - 40x + 3) is divided by 2 . Then (x^2+1)^4 is not simplified by any no. Lastly, why do u put -8? if u try to expand it back it didn't get the same ans.
Really. Forgotten your algebra I?
(8-2x)^2 = 4(4-x)^2
The intervening two lines should have had a "-" sign in front, from the (-2) factor. My typo. No need to pull out the 8; just felt like it. Prolly should leave it as 8-2x, so it looks more like the original syntax.
To differentiate the function f(x) = (8-2x)^3(x^2+1)^5, you can use the product rule and chain rule.
The product rule states that if you have two functions u(x) and v(x), the derivative of their product is given by:
(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x).
First, let's find the derivative of the first part of the function, (8-2x)^3. Applying the chain rule, we have:
u(x) = (8-2x), so u'(x) = -2.
Now, let's find the derivative of the second part of the function, (x^2+1)^5. Applying the chain rule again:
v(x) = (x^2+1), so v'(x) = 2x.
Applying the product rule, we get:
f'(x) = u'(x)v(x) + u(x)v'(x)
= (-2)(x^2+1)^5 + (8-2x)^3(2x).
Expanding further, we have:
f'(x) = -2(x^2+1)^5 + 2(8-2x)^3x
= -2(x^2+1)^5 + 2(8-2x)(8-2x)(8-2x)x.
Simplifying this expression results in:
f'(x) = -2(x^2+1)^5 + 2(8-2x)^3x
= -2(x^2+1)^5 + 2(64-32x+4x^2)x
= -2(x^10 + 5x^8 + 10x^6 + 10x^4 + 5x^2 + 1) + 2(64x - 32x^2 + 4x^3).
Expanding the expression further, we can simplify it to:
f'(x) = -2x^10 - 10x^8 - 20x^6 - 20x^4 - 10x^2 - 2 + 128x - 64x^2 + 8x^3
= -2x^10 + 8x^3 - 10x^8 - 64x^2 - 20x^6 - 20x^4 - 10x^2 + 128x - 2.
Hence, the derivative of f(x) is f'(x) = -2x^10 + 8x^3 - 10x^8 - 64x^2 - 20x^6 - 20x^4 - 10x^2 + 128x - 2.
Make sure to double-check your calculations and compare them with the final result provided above.