Derivative of y= x² tan(1/x)?

First, remember that:

d/dx[tan(x)] = sec²(x)

Product rule and chain rule (as you said):

y' = [x² * [sec²(1/x) * (-1/x²)]] + [tan(1/x) * 2x]

Simplify:

y' = [-1* sec²(1/x)] + [tan(1/x) * 2x]

y' = -sec²(1/x) + 2x tan(1/x)

Your calculator calculates the derivative of a function differently from the way a human would, so sometimes it gets the wrong answer for a derivative. For example, if you graph the function y=abs val (x) on your TI-84 and calculate the derivative at x=0, the calculator thinks the derivative is zero. However, the derivative of abs val (x) at x=0 is undefined because a derivative isn't defined at sharp peaks or valleys on a graph. Absolute value functions are shaped like a V, so there is no derivative at the sharp valley at x=0. Anyway, the point is that the calculator doesn't always get the right answer for a derivative. If you calculate the derivative with the Chain rule and the product rule, I think you get -sec^2(1/x)+2xtan(1/x).

First use the product rule.

d/dx[x²•tan(1/x)]=d/dx[x²]•tan(1/x) + d/dx[tan(1/x)]•x²

2x•tan(1/x) + d/dx[tan(1/x)]•x²

Now let's just pay attention to the derivative of tan(1/x). Replace 1/x with x^-1.

d/dx[tan(x^-1)]

The outer function is tan u, and the inner function is x^-1.

To apply the chain rule, multiply the derivative of the outer function (when u=x^-1) by the derivative of the inner function.

d/dx[tan u] • d/dx[x^-1]

Remember that the derivative of tan x is sec² x. (This can be derived through the quotient rule since tan x=sin x / cos x, but it's faster to memorize the derivatives of the trigonometric functions.)

Use the simple power rule to find the derivative of x^-1.

sec² u • -x^-2

Now replace u with 1/x. Simplify the expression -x^-2 so there are no negative exponents.

sec²(1/x) • (-1 / x²)

You can further simplify the expression by putting sec²(1/x) in terms of the cosine.

sec²(1/x) = 1 / cos²(1/x)

1 / cos²(1/x) • (-1 / x²)

(1 • -1) / (cos²(1/x) • x²)

-1 / x²cos²(1/x)

That's as far as I can solve it.

(tan(1x)(2x))(x^2)(sec^2(1/x))(-x^-2)

Sorry if its confusing. You do use the product rule and the chain rule. Trig works the same with chain run as anything else. The trig function becomes the outside part.

so its a chain and product rule.

take the derivative of X squared= 2x and multiply it time tan(1/x), that's the first part of the product rule. then take the derivative of tan(1/x) which you would need to take out the derivative of 1/x which is the ln(x) and multiply it times sec2(1/x), then multiply that by x squared.

so you'll get

2xtan(1/x)+X2ln(x)sec2(1/x)

use both the chain rule and the product rule.

y' = 2xtan(1/x) + (x^2) * sec^2(1/x) * (-1/(x^2))

the answer can then be simplified to

= 2xtan(1/x) + -sec^2(1/x)

Use the chain rule on the tan(1/x) portion and the product rule for the x^2tan(u) portion.

this is an example of derivative of the product of two function.

Let f be g*h (in your case g=x^2 and h=tan(1/x)).

f ' = g ' * h + g * h '

in your case

y ' = 2 x * tan(1/x) + x^2 *(1/ cos^2(1/x) * (-1/x^2))

I suggest buying a TI-89 calculator bc you can do so much with it and it will really help even with derivatives. And i would help but i forget all this math suff haha sorry

y= x² tan(1/x)

=2tan(1/x)*x-sec^2(1/x)

YEA GOOD JOB PIE FOR EVERYONE