I'm having some trouble differentiating radical functions

[Specter]

Homework Statement

Find the derivative of the following functions.

h(x)=8x²√x²+1

Homework Equations

The Attempt at a Solution

I just had a lesson on friday but it was a bit confusing to me. I am able to solve some problems but ones that contain multiple rules confuse me. If someone could break down or explain what is happening in this question to me that would be great. This question is solved in the lesson but I don't really know what is happening in each step.

h(x)=8x²√x²+1

h'(x)=8x²(x²+1)^1/2

So I understand that √x+1 is now written as x²+1^1/2

h'(x)=16x(x²+1)^1/2(8x²)1/2(x²+1)^-1/2(2x)

I understand that 8x² is now 16x (2*8x=16x). I know that the 1/2 is brought down and then the (x²+1)^1/2 becomes (x²+1)^-1/2 because (x²+1)^1/2-2/2. Next the deriative of x² is put down at the end but I don't really know why or what rule that is. From here on out I'm pretty lost on what is happening...

=16x(x²+1)^1/2+8x³/(x²+1)^1/2

=16x³+16x+8x³/(x²+1)^1/2

=24x³+16x/√x²+1

 

[fresh_42]

Homework Statement

Find the derivative of the following functions.

h(x)=8x²√x²+1

Homework Equations

The Attempt at a Solution

I just had a lesson on friday but it was a bit confusing to me. I am able to solve some problems but ones that contain multiple rules confuse me. If someone could break down or explain what is happening in this question to me that would be great. This question is solved in the lesson but I don't really know what is happening in each step.

h(x)=8x²√x²+1

h'(x)=8x²(x²+1)^1/2

So I understand that √x+1 is now written as x²+1^1/2

h'(x)=16x(x²+1)^1/2(8x²)1/2(x²+1)^-1/2(2x)

I understand that 8x² is now 16x (2*8x=16x). I know that the 1/2 is brought down and then the (x²+1)^1/2 becomes (x²+1)^-1/2 because (x²+1)^1/2-2/2. Next the deriative of x² is put down at the end but I don't really know why or what rule that is. From here on out I'm pretty lost on what is happening...

=16x(x²+1)^1/2+8x³/(x²+1)^1/2

=16x³+16x+8x³/(x²+1)^1/2

=24x³+16x/√x²+1

Do you know the product rule and chain rule of differentiation? They are useful to deal with composite functions by breaking them down in easier ones: $(f(x)\cdot g(x))'$ and $f(g(x))'$.

 

[Specter]

Do you know the product rule and chain rule of differentiation? They are useful to deal with composite functions by breaking them down in easier ones: $(f(x)\cdot g(x))'$ and $f(g(x))'$.

I do know them. The chain rule in my notes is written as, if F(x)=(f(x))ⁿ then F'(x)=nf'(x)f(x)^n-1 and the product rule is written as F'(x)=f'(x)g(x)+f(x)g'(x). If I know both of these rules, is there a specific order to use when solving the question? How do I know what to do first if I have to use multiple rules? I'm taking gr 12 calculus online so if I have questions like this I never really get a chance to ask, which is why so many things I should know by now still confuse me.

 

[fresh_42]

I do know them. The chain rule in my notes is written as, if F(x)=(f(x))ⁿ then F'(x)=nf'(x)f(x)^n-1 and the product rule is written as F'(x)=f'(x)g(x)+f(x)g'(x). If I know both of these rules, is there a specific order to use when solving the question? How do I know what to do first if I have to use multiple rules? I'm taking gr 12 calculus online so if I have questions like this I never really get a chance to ask, which is why so many things I should know by now still confuse me.

Just break it down:
$$
h(x)= 8x^2\sqrt{x^2+1} = 8 \cdot x^2 \cdot \left(\sqrt{(x^2+1)}\right)
$$
so you have a constant factor as one rule, then a product, and at last the chain rule for the differentiation of the root factor. The order is given by the function: $h(x)=c \cdot f(x)\cdot g(k(x))$ so $h' = c \cdot (f \cdot g)'$ and so on.

 

[Specter]

Just break it down:
$$
h(x)= 8x^2\sqrt{x^2+1} = 8 \cdot x^2 \cdot \left(\sqrt{(x^2+1)}\right)
$$
so you have a constant factor as one rule, then a product, and at last the chain rule for the differentiation of the root factor. The order is given by the function: $h(x)=c \cdot f(x)\cdot g(k(x))$ so $h' = c \cdot (f \cdot g)'$ and so on.

Okay that helps me out a bit. I tried on paper and this is as far as I get.

So the rules I am using are...

Constant multiple rule, product rule, and the chain rule.

f(x)=8x²
f'(x)=16x

g(x)=(x²+1)^1/2
g'(x)=1/2(x²+1)^-1/2

h(x)=8x²√x²+1
h(x)=8x²(x²+1)^1/2h'(x)=16x(x²+1)^1/2+(8x²)1/2(x²+1)^-1/2

I followed the product rule but this step isn't correct according to the solution in the lesson. I'm not sure what to do next.

 

[Mark44]

Find the derivative of the following functions.
h(x)=8x²√x²+1

Please use parentheses to make what you write clearer.
This --- √x²+1 -- should be written as √(x²+1), otherwise it looks like $\sqrt{x^2} + 1$, which isn't what you intended.

At the bottom of your post you have =24x³+16x/√x²+1
You need parentheses here as well -- =(24x³+16x)/√(x²+1) -- I think this is what you meant, but I can't tell.

New members here aren't used to having to write fractions on one line. If someone writes a + b/c + d, with the intention of writing $\frac{a + b}{c + d}$, that's not how what was written would be interpreted. Instead, it would be seen as $a + \frac b c + d$, a long way from what they really meant.

 

[fresh_42]

f(x)=8x²
f'(x)=16x

O.k. although it is a bit strange to actually break $h$ apart. It is more like $h(x)=8\cdot A(x)$ and $h'(x)=8 \cdot A'(x)$. I used $A$ in order not to confuse it with your notation. Here we have $A(x)=x^2\sqrt{x^2+1}$ and we are left with the task to calculate $A'(x)$.

g(x)=(x²+1)^1/2
g'(x)=1/2(x²+1)^-1/2

Here you've forgotten the chain rule. $g(x)=\sqrt{x^2+1}=\sqrt{k(x)}= \sqrt{} (\, k(x) \,)$ and $g'(x)=\sqrt{k(x)}' \cdot k'(x)$ where $k(x)=x^2+1$. You differentiated the square root, but not the nested function $k(x)$ which you will have to. The reason is the following: Say we want to differentiate $\sqrt{x^2+1}=\sqrt{k(x)}$. Now to make it easy, say $y=k(x)$, so the task is to find $\sqrt{y}'$ which is $\frac{1}{2}y^{-\frac{1}{2}}$ but only if we differentiate it with $y$ as variable. However, we want to differentiate with $x$ as variable. So we computed actually $\frac{d \sqrt{y}}{dy}$ and not $\frac{d \sqrt{y}}{dx}$. To correct this, we have to multiply by $\frac{dy}{dx}=\frac{d k(x)}{dx }=\frac{d(x^2+1)}{dx}$. The differentiation w.r.t. $y$ has to be turned into one w.r.t. $x$ and $y'(x)=k(x)$ does this.

h(x)=8x²√x²+1
h(x)=8x²(x²+1)^1/2
h'(x)=16x(x²+1)^1/2+(8x²)1/2(x²+1)^-1/2
I followed the product rule but this step isn't correct according to the solution in the lesson. I'm not sure what to do next.

As said, your correction from the chain rule is missing. What to do next? I suggest, after the correction, to multiply all constant factors, and maybe write the entire thing as one quotient, because $(x^2+1)^{-\frac{1}{2}}=\frac{1}{\sqrt{x^2+1}}$.