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

[diffusion]

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

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

Can someone please correct me if I have made any mistakes?

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?

 

[psycho2499]

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.

 

[Russell Berty]

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

 

[diffusion]

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.

 

[Russell Berty]

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.

 

[psycho2499]

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 into 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

 

[Russell Berty]

Questions about the previous post:

? 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.

 

[psycho2499]

Questions about the previous post:

? 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.

It would be the chain rule but in tandum with the product rule which I forgot to do anyway, and thank you for catching that.