can please someone help me how to solve for the derivative of these
y = [t^2 + (1/t)]^2
G(x) = {1+[x+(x^2+x^3)^4]^5}^6
h(x) = {x+[x+(x)^1/2]^1/2}^1/2
y = x^1/2 sin [x+(x)^1/2]^1/2 cot x^1/2
if possible, not simplified. thanks
$\displaystyle y = (t^2 + (1/t))^2$, firstly rewrite this in the form $\displaystyle y = (t^2 + t^{-1})^2)$.
To apply the chain rule you take the function inside the brackets to the front, like so:
$\displaystyle = 2t - t^{-2} . (t^2 + t^{-1})^2$
You then multiply the whole thing by the power of the bracket:
$\displaystyle = 2t - t^{-2} . (t^2 + t^{-1})^{2} . 2$
Then finally you reduce the power of the bracket by one, ending up with:
$\displaystyle \frac{dy}{dx}= 2t - t^{-2} . (t^2 + t^{-1})^{1} . 2$
$\displaystyle \frac{dy}{dx} = 2(2t - t^{-2})(t^2 + t^{-1})$
Im sure you can simply the remaining equation and apply the same rules to your other questions
The second and third steps are incorrect. Here's a way that always work.
First let me say that the chain rule is used for a composition of functions. In this case, the order of operation is
(i) $\displaystyle t^2 + \frac{1}{t}$
(ii) then square the result from (i) i.e. $\displaystyle \left( t^2 + \frac{1}{t} \right)^2$
For the chain rule, let
$\displaystyle u = t^2 + \frac{1}{t}$
so
$\displaystyle y = u^2$
Then the derivative $\displaystyle \frac{dy}{dt}$ is
$\displaystyle \frac{dy}{dt} = \frac{dy}{du} . \frac{du}{dt}$
so $\displaystyle \frac{dy}{du} = 2u$ and $\displaystyle \frac{du}{dt} = 2t - \frac{1}{t^2}$
so
$\displaystyle \frac{dy}{dt} = \frac{dy}{du} . \frac{du}{dt} = 2u \left( 2t - \frac{1}{t^2} \right) = 2 \left( t^2 + \frac{1}{t} \right) \left( 2t - \frac{1}{t^2} \right) $
It is the answer given above but I think it's necessary to put in these steps until you get the hang of it. Once you do, you can write the answer in one step (working from out to in on your function).