# What is the derivative of (tan(x))^(1/x) and sqrt((x-1)(x^2 (x-4)))?

## Homework Statement

Find the following derivatives:

$$\frac{d}{dx}$$ (tanx)^(1/x)

$$\frac{d}{dx}$$ $$\sqrt{\frac{x-1}{x^2 (x-4)}}$$

## Homework Equations

Rules of differentiation

## The Attempt at a Solution

I realize these are fairly elementary problems, just not sure the order I should perform the differentiation rules. My attempt at number 1 is as follows:

Considering 1/x as a separate function (u), we can use the chain rule:

$$\frac{d}{dx}$$tan(x)$$_{u}$$ $$\frac{d}{dx}$$ $$\frac{1}{x}$$

Apply quotient rule on 1/x:

= sec$$_{2}$$ (x)$$_{u}$$ $$\frac{1}{x^2}$$

= $$\frac{sec^2(x) ^{1/x}}{x^2}$$

The second problem I'm not sure. As far as I can tell it's just applying the quotient rule to the fraction, all to the power of $$\frac{1}{2}$$. Do I need to apply the chain rule to the denominator before using the quotient rule?

Well the only mistake I see in your first work is that you forgot a negative. and for the second problem break it down to a more simple problem. It may become more work to resimplify it later but you aren't wrong to do so.

You can take it and make it into this d/dx((x-1)(1/2)/(x2(x-4))(1/2)) which can then become d/dx((x-1)(1/2)*(x3-4x2)(-1/2)). Which is much easier to differentiate by the product rule or course you can always stop one step short of this and use the quotient rule.

To do the first one, you will either need to use logarithmic differentiation, or rewrite the expression as

e^ln((tanx)^(1/x)) which is e^((1/x)(ln(tanx)))

Now the base is a constant, in fact e, so it looks like e^u... use the chain rule now

Well the only mistake I see in your first work is that you forgot a negative. and for the second problem break it down to a more simple problem. It may become more work to resimplify it later but you aren't wrong to do so.

Missing a negative where?

To do the first one, you will either need to use logarithmic differentiation, or rewrite the expression as

e^ln((tanx)^(1/x)) which is e^((1/x)(ln(tanx)))

Now the base is a constant, in fact e, so it looks like e^u... use the chain rule now

You sure about that one? Just seems like a lot of work for such a simple function.

Consider the following function: x^x

Find (d/dx)x^x

You cannot use either of the rules (d/dx)x^n = nx^(n-1) or (d/dx)e^x = e^x.
In the first rule the exponent is a CONSTANT, in the second the base is a CONSTANT (in particular it is e.)
You also cannot use the more general version of the second rule, (d/dx)a^x = ln(a)*a^x because the base is a CONSTANT.

In the expression x^x, neither the base nor the exponent are constant.

So you use logarithmic differentiation, or you rewrite it as e^(ln(x^x)) which is e^(x*lnx) then, let u = x*lnx and it looks like e^u, at which point you can use the chain rule along with the fact that (d/dx)e^x = e^x.

Sorry this took so long. Originally I was looking at how you did it not the question and what the answer should be. You forgot a negative from d/dx(1/x) but I'm pretty sure you can't do it that way. As far as I know you can approach this straight on or using a ln trick. I'll show both.

Straight on
1)Take derivative of exponent and put at front then subtract 1 from part with exponent then tack on the derivative of the inside on the end. Like so:
d/dx(1/x)(tan(x))^((1/x)-1)d/dx(tan(x))
2)Simplify
-((tan(x)^((1-x)/x)sec(x)^2)/x^2)

Or with ln

1) set it equal to y
y=tan(x)^(1/x)
2)ln both sides
ln(y)=ln(tan(x)^(1/x))
3)move (1/x) to front in accordance with laws of ln
ln(y)=(1/x)ln(tan(x))
4)differentiate
dy/y=(-1/x^2)(1/tan(x))(sec(x)^2)
5)times by y on either side and sub back in to eliminate it since y=tan(x)^(1/x)
dy=(-1/x^2)(tan(x)^(-1))(sec(x)^2)(tan(x)^(1/x))
6)simplify
dy=-(sec(x)^2*tan(x)^((1-x)/x))/x^2

??? d/dx(1/x)(tan(x))^((1/x)-1)d/dx(tan(x)) What rule is being used?

You cannot use d/dx x^n = n*x^(n-1) here (with or without a chain rule.)

??? ln(y)=(1/x)ln(tan(x)) ===> dy/y=(-1/x^2)(1/tan(x))(sec(x)^2) You need to use the product rule here.