The right derivative of f= (sin x)^(cos x) +(cos x)^(sin x) when x=0

[Crivi]

How do you compute the value of the right derivative of f= (sin x)^(cos x) +(cos x)^(sin x) when x=0. I'm trying to learn calculus so some explanations wouldn't be so bad.I got stucked computing the limit of (sin x)^(cos x) *( (cos ^2 x)/(sin x) - sin (x) * ln (sin x) ), x->0. Sorry for the grammar mistakes but i'm not english.

 

[Deleted member 4993]

How do you compute the value of the right derivative of f= (sin x)^(cos x) +(cos x)^(sin x) when x=0. I'm trying to learn calculus so some explanations wouldn't be so bad.I got stucked computing the limit of (sin x)^(cos x) *( (cos ^2 x)/(sin x) - sin (x) * ln (sin x) ), x->0. Sorry for the grammar mistakes but i'm not english.

I cried uncle and went wolframalpha.com to find:

 

[Steven G]

I thought that Sal cried uncle when he couldn't do a problem? I guess you have to cry to someone too.

 

[Steven G]

Here is how I would do such a problem.
f(x) = sin(x))^(cos(x) + cos(x))^(sin(x)

So f'(x) = der[(sin(x))^(cos(x)]+ der[cos(x)^sin(x)]

I would calculate those derivatives separately!

Let y = sin(x))^(cos(x)

Then ln(y) = cos(x)*ln(sin(x))

Now take the derivative of both sides wrt x.

y'/y = -sin(x)*ln(sin(x)) + cos(x)*cos(x)/sin(x)

Now y' = y[-sin(x)*ln(sin(x)) + cos(x)*cos(x)/sin(x)] = sin(x)^cos(x)[-sin(x)*ln(sin(x)) + cos(x)*cos(x)/sin(x)]

Now let y = (cos(x))^(sin(x)) and compute y'

Finish up!

 

[Steven G]

Here is how I would do such a problem.
f(x) = sin(x))^(cos(x) + cos(x))^(sin(x)

So f'(x) = der[(sin(x))^(cos(x)]+ der[cos(x)^sin(x)]

I would calculate those derivatives separately!

Let y = sin(x))^(cos(x)

Then ln(y) = cos(x)*ln(sin(x))

Now take the derivative of both sides wrt x.

y'/y = -sin(x)*ln(sin(x)) + cos(x)*cos(x)/sin(x)

Now y' = y[-sin(x)*ln(sin(x)) + cos(x)*cos(x)/sin(x)] = sin(x)^cos(x)[-sin(x)*ln(sin(x)) + cos(x)*cos(x)/sin(x)]

Now let y = (cos(x))^(sin(x)) and compute y'

Finish up!

Subhotosh, I appreciate your approval. Thank you!

 

[Dr.Peterson]

I think the OP's difficulty is not in differentiating (though only half the derivative is shown), but in evaluating it at x=0 (from the right), that is, finding the limit as [MATH]x\rightarrow 0^+[/MATH].

I'd like to know why someone who is "trying to learn calculus" would be given something this ugly. Is there some context to this problem? Is it part of something bigger?