Thread: Differentiate Y = 5 ^ (-1/x)

1. Differentiate Y = 5 ^ (-1/x)

The problem is as follows: Differentiate Y = 5^(-1/x)

My solution:

since d/dx(a^x) = a^x * ln(a)
I got dy/dx to be 5^(-1/x)ln5.

this is incorrect, the answer should be [5^(-1/x)ln5] / x^2

Thus, my question is where does that x ^ 2 denominator come from?

2. Re: Differentiate Y = 5 ^ (-1/x)

Originally Posted by iAmKrizzle
The problem is as follows: Differentiate Y = 5^(-1/x)
this is incorrect, the answer should be [5^(-1/x)ln5] / x^2
Implicitly differentiate $\ln(y)=\frac{\ln(5)}{x}$ and solve for $y'$.

3. Re: Differentiate Y = 5 ^ (-1/x)

While the implicit differentiation seems easier, this problem is in a section before implicit differentiation is covered. This problem appears in a section on the chain rule, how would I solve it without using implicit differentiation?

4. Re: Differentiate Y = 5 ^ (-1/x)

Originally Posted by iAmKrizzle
While the implicit differentiation seems easier, this problem is in a section before implicit differentiation is covered. This problem appears in a section on the chain rule, how would I solve it without using implicit differentiation?
Well frankly there is really no other way to do this sort of problem.
The solution is based on this theorem:
If $f(x)$ has a derivative and $c>0$ then if $y=c^{f(x)}$ then $y'=f'(x)c^{f(x)}\ln(c)$.
That theorem is proved by implicit differentiation.

5. Re: Differentiate Y = 5 ^ (-1/x)

Understood. Thank you I will apply this principle.

6. Re: Differentiate Y = 5 ^ (-1/x)

Originally Posted by Plato
Well frankly there is really no other way to do this sort of problem.
The solution is based on this theorem:
If $f(x)$ has a derivative and $c>0$ then if $y=c^{f(x)}$ then $y'=f'(x)c^{f(x)}\ln(c)$.
That theorem is proved by implicit differentiation.
It appears the OP is expected to know how to change the base of an exponential.

$5^{-1/x} = e^{\log \left(5^{-1/x}\right)} = e^{\frac{-1}{x} \log(5)}\right)} = e^{\frac{k}{x}}$ where $k = -\log(5)$.

After differentiating, the answer ought to be re-written in base 5 ....