Question

differentiate f(x)= -2x/sqrt(3-7x)
i got (sqrt(3x-7)(-2)-(-2x)(1/2)(3x-7)^-1/2))/(sqrt(3x-7)^2) but that is incorrect..any insight?

Answer 1

\[f(x)=\frac{ -2x }{ \sqrt{3-7x}}\]
\[f'(x)=\frac{ (-2)(\sqrt{3-7x})-(-2x)((0-7(1))(1/2)(3-7x)^{1/2-1} }{ (\sqrt{3-7x})^{2} }\]

Answer 2

you just need to simplify it after

Answer 3

is everything I did clear?

Answer 4

yessir thank you!

Answer 5

\[f'(x)=\frac{ -2 }{ \sqrt{3-7x} }-\frac{ 7x }{ (\sqrt{3-7x})^{3} }\]

Answer 6

no problem!

Answer 7

another one if you dont mind du/dx sqrt(2+sin^2(x)) i got
1/2(2+sin^2(x))^(-1/2) but that is incorrect, is it correct when applying the chain rule you leave the inside equation totally alone?

Answer 8

no, the chain rule is
\[(k)*(f'(x))*(f(x))^{k-1}\]

Answer 9

you're missing f'(x) in your answer. And with this function, f'(x) will require you to use the chain rule again. Inception!

Answer 10

yeah i always get confused and forget to take the derivative of whats inside... on another one du/dx (x^6)(e^5x) i tried
(e^5x)(6x^5)+(x^6)(e^5x) to no avail....... isnt du/dx e^5x =e^5x?

Answer 11

no, it's 5e^5x.
You need to derive the exponent

Answer 12

awesome, thanks again man !

Answer 13

no problem!