Question

What is the derivative of y= 5^(-1/x)? The answer says
y= 5^(-1/x)
= [5^(-1/x)][In5][(-1)(-x^(-2)]
= [5^(-1/x)(In5)]/x^2
but I don't get the second step "-x^(-2)".. why is x a negative, not positive? Since the exponent of the original question is (-1/x), shouldn't the inner derivative be [(-1)(x^(-2)] ?

Answer 1

y = 5^(-1/x) implies that
ln y = -1/x . ln 5

=> y'/y = 1/x^2 . ln 5

=> y' = ln 5 . 1/x^2 . y
= ln 5 . 1/x^2 . 5^(-1/x)

Answer 2

Sorry, but can you put it in simpler terms (if possible)? We are learning about the chain rule and how to apply them...

Answer 3

Write y =5^u where u = -1/x

Then dy/dx = dy/du . du/dx

Now du/dx = 1/x^2, yes?
and
dy/du = ln 5 . 5^u

so far so good?

Answer 4

Yes, I'm following

Answer 5

Then

dy/dx
= dy/du . du/dx
= ln 5 . 5^u . (1/x^2)
= ln 5 . 5^(-1/x). 1/x^2

Answer 6

Thank you! But I was wondering, if this is correct? u=-1/x = = -x^(-1). I can't seem to differentiate -x^(-1). I get dy/dx= (-1).(x^-2)

Answer 7

If f(x) = 1/x = x^(-1), then df/dx = (-1)x^(-1-1) = -1/x^2

hence for our function u = -1/x, du/dx = -(-1/x^2) = 1/x^2

Answer 8

Oh okay. Thanks so much!