The problem is: tansqrt(5t). I get that I need to use the chain rule but I think I'm messing it up.

So I did this: tansqrt(5t) = sec^2sqrt(5t) * 1/2(tt)^-1/2 = sec^2sqrt(5t)/2sqrt(5t).

However, this is the wrong answer. What am I missing?

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- Mar 20th 2013, 08:29 PMAlucard2487Can anyone explain this problem to me?
The problem is: tansqrt(5t). I get that I need to use the chain rule but I think I'm messing it up.

So I did this: tansqrt(5t) = sec^2sqrt(5t) * 1/2(tt)^-1/2 = sec^2sqrt(5t)/2sqrt(5t).

However, this is the wrong answer. What am I missing? - Mar 20th 2013, 08:43 PMibduttRe: Can anyone explain this problem to me?
In the end multiply by the derivative of 5t i.e., 5. Rest is fine except one typing error, for 1/2(tt)^-1/2 It should be 1/2(5t)^-1/2

- Mar 20th 2013, 09:46 PMAlucard2487Re: Can anyone explain this problem to me?
I see what I did wrong. I forgot to do the chain rule for the inside.

- Mar 21st 2013, 12:20 AMProve ItRe: Can anyone explain this problem to me?
I always use Leibnitz notation when using the Chain Rule. You can have as many links in your chain as you need. Here we can see the 5t function inside a square root function inside a tangent function. Since it's three functions deep, we need three links in the chain.

$\displaystyle \displaystyle \begin{align*} u &= 5t \implies y = \tan{ \left( \sqrt{u} \right) } \\ \\ v &= \sqrt{u} \implies y = \tan{(v)} \\ \\ \frac{du}{dx} &= 5 \\ \\ \frac{dv}{du} &= -\frac{1}{2\sqrt{u}} = -\frac{1}{2\sqrt{ 5t }} \\ \\ \frac{dy}{dv} &= \sec^2{(v)} = \sec^2{ \left( \sqrt{u} \right) } = \sec^2{ \left( \sqrt{ 5t } \right) } \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dv}{du} \cdot \frac{dy}{dv} \\ &= 5 \left( -\frac{1}{2\sqrt{ 5t} } \right) \sec^2{ \left( \sqrt{5t} \right) } \\ &= -\frac{5\sec^2{ \left( \sqrt{5t} \right) }}{2\sqrt{5t}} \end{align*}$ - Mar 21st 2013, 06:42 AMAlucard2487Re: Can anyone explain this problem to me?
I'm a bit confused on the way you did this. Also, when I typed the correct answer in, I got a positive answer.

I ended up with (5sec^2sqrt(5t)/2sqrt(5t)) * (dt). - Mar 21st 2013, 02:32 PMProve ItRe: Can anyone explain this problem to me?
Yes you are correct, my dv/du should be positive, giving a positive answer.

I don't know where you're pulling *dt from though... - Mar 21st 2013, 04:14 PMAlucard2487Re: Can anyone explain this problem to me?
My apologies. The question was to find the differential of each function. That's where the dt came from.