Product rule

[Mel Mitch]

Please help with my product rule


y= x^4(x+3)^5

My steps :

u= x^4
du/dx= 4x^3

v=(x+3)^5
dv/dx= 5(x+3)^4 X 1

using
dy/dx= v x du/dx + u x dv/dx

(x +3)^5 x 4x^3 + x^4 x x^4 x 5(5 +3)^4

4x^3(x+3)^5 + 5x^4 (x +3)^4



some part of this is wrong but i'm not sure.....so i'm unable to complete

 

[arthur ohlsten]

first times derivative of the second PLUS the second times derivative of the first

y=x^4[x+3]^5
dy/dx= x^4 [5(x+3)^4] + [x+3]^5 4x^3

u=x^4
du/dx=4x^3

v=[x+3]^5
dv/dx=5[x+3]^4 d/dx[x+3]
dv/dx=5[x+3]^4 [1]

Arthur

 

[Mel Mitch]

i am not getting the answer....

 

[BigGlenntheHeavy]

\(\displaystyle Let \ f(x) \ = \ y \ = \ x^{4}(x+3)^{5}, \ then \ f '(x) \ = \ y' \ = \ x^{4}[5(x+3)^{4}] +4x^{3}(x+3)^{5}.\)

\(\displaystyle = \ (x+3)^{4}[5x^{4}+4x^{3}(x+3)] \ = \ (x+3)^{4}[5x^{4}+4x^{4}+12x^{3}]\)

\(\displaystyle = \ (x+3)^{4}(9x^{4}+12x^{3}) \ = \ 9x^{8}+120x^{7}+630x^{6}+1620x^{5}+2025x^{4}+972x^{3}.\)

 

[Mel Mitch]

thank you so much bigGlentheheavy..........i now know my error.....

 

[daon]

Another way:

\(\displaystyle x^4 = (x^{\frac{4}{5}})^5 \implies x^4(x+3)^5 = (x^{\frac{9}{5}} + 3x^{\frac{4}{5}})^5\)

So,

\(\displaystyle f'(x) = 5(x^{\frac{9}{5}} + 3x^{\frac{4}{5}})^4 \cdot (\frac{9}{5}x^\frac{4}{5} + \frac{12}{5}x^{-\frac{1}{5}})\)

 

[Mel Mitch]

Thank you daon......i'm always looking for another way....thanks again