Question

Differentiate. f(x) = 5 x sin x

Answer 1

\[5\sqrt{x}\sin x\]

Answer 2

use the product rule

let f(x) = sqrt(x) and g(x) = sin(x)

Answer 3

the 5 is a coefficient which can be temporarily pulled out, and ignored (but don't forget to place it back in when you're done deriving)

Answer 4

ok thats what i was doing wrong. Forgot to put the 5 back in. Thank you

Answer 5

you're welcome

Answer 6

Find the derivative of the function. y=3^(8-x^2)

Answer 7

you will use the exponential rule

http://www.mathsisfun.com/calculus/derivatives-rules.html

and the chain rule

Answer 8

when you get to 8-x^2, you'll use the power rule

Answer 9

tell me what you get

Answer 10

-2x(ln3)3^(8-x^2)

Answer 11

does thae same rule apply to base g? or is that a constant? as in \[g^{\sin pix}\]

Answer 12

oh never mind it actually says 9 not g.

Answer 13

-2x(ln3)3^(8-x^2) is correct

Answer 14

and if they say "derive with respect to x", then the other letters are usually constants (unless they state something like "g is a function of x")

Answer 15

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 50 cm/s. Find the rate at which the area within the circle is increasing after 3 seconds. I get 47,123 instead of 15000pi. What am i missing

Answer 16

"A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 50 cm/s". So this means that the radius is growing at a speed of 50 cm/s.


So let's first find the radius function (function of time)


Length of Radius = (Speed)*(Time)


r = s*t


\[\Large r = 50t\]


Then derive this radius function with respect to t. This is effectively the same as its speed.


\[\Large \frac{dr}{dt} = 50\]


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Plug t = 3 into the equation r = 50t to get the radius at time t = 3


\[\Large r = 50t\]


\[\Large r = 50(3)\]


\[\Large r = 150\]


At time t = 3, the radius is 150 cm.


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Now derive the area function with respect to t and plug in the values found above


\[\Large A = \pi*r^2\]


\[\Large \frac{dA}{dt} = 2\pi*r*\frac{dr}{dt}\]


\[\Large \frac{dA}{dt} = 2\pi*150*50\]


\[\Large \frac{dA}{dt} = (2*150*50)*\pi\]


\[\Large \frac{dA}{dt} = 15,000\pi\]