Question

Just looking for someone to check if my answer is right.

Calculate the derivative:
y=sqrt2 +2/x
at yprime(16)

I had 1/2

Answer 1

\[y=\frac{ \sqrt{x}+2 }{ x }\]

Answer 2

\[y ^{'}(16)\]

Answer 3

What did you get for y ' ?

Answer 4

i think i got it wrong, because it's pretty wonky.
\[\frac{ x \sqrt{x}-4\sqrt{x} }{ 2\sqrt{x} } \times \frac{ 1 }{ x ^{2} }\]

Answer 5

i used the quotient rule -
\[\frac{ x \times \frac{ 1 }{ 2\sqrt{x} }-\sqrt{x}+2(1) }{ x ^{2} }\]

Answer 6

I like the product rule (it's easier to remember in my opinion)

y = (sqrt(x) + 2)/x

y = (sqrt(x) + 2)*x^(-1)

y' = d/dx[(sqrt(x) + 2)]*x^(-1) + (sqrt(x) + 2)*d/dx[x^(-1)]

y' = 1/(2*sqrt(x))*x^(-1) + (sqrt(x) + 2)*(-x^(-2))


You can simplify a number of ways, but you can jump to the part where you plug in x = 16 and evaluate to get...


y'(16) = 1/(2*sqrt(16))*(16)^(-1) + (sqrt(16) + 2)*(-(16)^(-2))

y'(16) = -1/64 .... or .... y'(16) = -0.015625

Answer 7

how can i use the product rule when it's division?

Answer 8

the quotient x/y can be rewritten as a product x*y^(-1)

Answer 9

since 1/y = y^(-1)

Answer 10

so i change it into a product then i can just do the product rule,
can i only do this when there is just one variable in the denominator?

Answer 11

you can do it for any algebraic expression in the denominator

A more complicated example

\[\Large \frac{x^2+3x}{3x^3+9x^2+7x}\]

is the same as

\[\Large (x^2+3x)(3x^3+9x^2+7x)^{-1}\]

Answer 12

You just write the numerator and denominator as a product and you write an exponent of -1 over the second part

Answer 13

i see!

Answer 14

of course, the quotient rule still works (I just forget it all the time and have to look it up constantly...so I rewrite the quotient into a product and use the product rule since it's easier to remember for me)