Some help with differentiation, please? (screenshots)

I have two separate homework questions I'm having issues with.

First Question:

Applying product rule:

f'(x) = 2(1+x)(e^-x) + (-e^-x)((1+x)^2)

Simplifies as:

f'(x) = (-e^-x)(x^2-1)

It's here that I don't know what to do. How would I go about the rest of the question?












Second question:

First of all, is that two separate functions? i.e. - Would I be using the product rule to differentiate? Or is it the log *of* the square root of x^2+3?

Also, how would you do the second part?



Thanks guys.

1) Find where f'(x)>0 as that will give you the range of values for which the function is increasing.

How would I do that?

(-e^-x)(x^2-1) > 0

What is it I do to find what the x values are?

Sorry, I'm rusty as hell after the holidays. No maths for 2-3 months.

Haha, me too.

split it into 2 parts

so find where:

-e^-x > 0
and
x^2-1 > 0

You should run into a problem with the first one.

No idea what to do here?

Solve the inequalities for x. You do understand why you do this, right?

Pretty sure I did at some point, lol.

I honestly can't remember learning anything about increasing/decreasing functions. All I remember doing is stationary points.

No idea what to do here?

So

-e^-x > 0
you can't solve this for x as ln(0) is undefined

x^2-1 > 0
x^2 > 1
mod(x) > 1

(or x > 1 and x < -1)

What you need to know for this is that the derivative finds the gradient and when the gradient is greater than 0 a function is increasing.

Ahh, yeah, I remember now. You could find if the graph of the function was negative/positive or decreasing/increasing by plugging x values into the function's derivative. We did that when finding the nature of stationary points.

So, my answer would be that the function is increasing when x is greater than 1 and when x is less than 1?

Yes

Yes

Your derivative is wrong because of a mistake in your working shown above in red.

$\displaystyle 2(1+x)e^{-x}-e^{-x}(1+x)^2 = e^{-x}(1+x)\left(2-(1+x)\right)$

$\displaystyle = e^{-x}(1+x)\left(1+x)\right) = e^{-x}(1+x)^2$

so where do i go from there?

e^-x > 0

still comes out as undefined, does it not?

so where do i go from there?

e^-x > 0

still comes out as undefined, does it not?

..

Your derivative was correct. So your problem boiled down to solving this inequality

$-e^{-x}(x^2-1)>0$

Firstly $e^{-x}$ is always positive which means that $-e^{-x}$ is always negative. So for the above inequality to hold, we need $x^2-1$ to also be negative i.e. we need to solve

$x^2-1<0$

You could have written your derivative as $e^{-x}(1-x^2)$ and then since $e^{-x}$ is always positive, you would need to solve

$1-x^2>0$

which is equivalent to the previous inequality.

Does this make sense and are you OK solving the quadratic inequality?

I get what you're saying.

As for the quadratic inequality... would it be:

1 - x^2 > 0
- x^2 > -1
x^2 > 1
x > -1, 1

I don't really know what to deduce from that it terms of an increasing function...

I don't really know what to deduce from that it terms of an increasing function...