what is the derivative of the square root of 5 over x to the 7th power?

Answer 1

y = root(5) / x^7 = 5^(½) x^(-7)

y' = (-7)• 5^(½)•x^(-8)
y' = -15.65 / x^8

Negative 15.65 over x to the 8.

Answer 2

the respond is 7/2?(7x) You ignored the 7 decrease than the unconventional. rather worth remembering that if y = ?x then dy/dx = one million/(2?x) So on your problem use that with the chain rule, differentiating the 7x to get the 7 on marvelous. in case you want to continually derive it with fraciotnal powers, then f(x) = (7x)^½, so f '(x) = ½ (7x)^-½•7 = 7/(2?(7x)) like I first suggested.

Answer 3

If you mean y = sqrt(5/x) then y = sqrt(5).x^(-1/2)

sqrt(5) is a constant and you must find the derivative of x^(-1/2)

y´(x) = sqrt(5)(-1/2)x^(-3/2)

y´´(x) = sqrt(5)(3/4)x^(-5/2)

y´´´(x) = sqrt(5)(-15/8)x^(-7/2)

.... I am sure you are able to reach there.

Answer 4

f(x) = [sqrt(5)] / [x^7]

f'(x) = [0 - sqrt(5)*7x^6] / [x^14]

f'(x) = [-sqrt(5)*7x^6] / [x^14]

f'(x) = [-7sqrt(5)] / [x^8]

If you meant f(x) = [sqrt(5/x)]^7, then it's a little more complicated, but not really that hard. You will need to use the chain rule twice. I'm too tired to do that one for you, but can look up the chain rule in your calc. book if you need help.

Answer 5

it is -7*(sqrt5) / x to the power 8

Answer 6

Who wants to know???

Answer 7

see
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=mathcom&s1=calculus&s2=differentiate&s3=basic

Answer 8

c.