Urgent help please -- Differentiation technique question

[Parsifal1]

I'm revising differentiation for tomorrow, I am hopelessly stuck on some things.

I have some examples with answers, but I have no idea which rules are being used.

y=2sin3x dy/dx=6cos3x

I see that sin goes to cos but why is the two multiplied by the 3 (if that's what's happening)?

y=3sin1/2x+e^x dy/dx= 6cos1/2x+e^x

Thank you

 

[fresh_42]

It is the chain rule and in your second equation there might be something wrong.

 

[pasmith]

I'm revising differentiation for tomorrow, I am hopelessly stuck on some things.

I have some examples with answers, but I have no idea which rules are being used.

y=2sin3x dy/dx=6cos3x

I see that sin goes to cos but why is the two multiplied by the 3 (if that's what's happening)?

$$\frac{d}{dx} \sin kx = k \cos kx$$ for constant $k$, as can be established by application of the chain rule.

I wish you luck, but I fear you have left your revision too late.

 

[Parsifal1]

$$\frac{d}{dx} \sin kx = k \cos kx$$ for constant $k$, as can be established by application of the chain rule.

I wish you luck, but I fear you have left your revision too late.

Okay thank you for the help, I have one more question which seems to need the chain rule, although I have an answer for it which looks false

Here is the question and the answer I was given:

y=3sin1/2x dy/dy=6cos1/2x

And the answer I got with chain rule:

1.5cos1/2x

 

[Mark44]

Okay thank you for the help, I have one more question which seems to need the chain rule, although I have an answer for it which looks false

Here is the question and the answer I was given:

y=3sin1/2x dy/dy=6cos1/2x

And the answer I got with chain rule:

1.5cos1/2x

Your answer is correct, if I can interpret what you think you are writing.

Please use parentheses!
Without parentheses, any of these are possible interpretations of the cosine part.
$\frac{\cos(1)}{2x}$
$\frac{\cos(1)}{2}x$
$\cos(\frac 1 2) x$
$\cos(\frac 1 {2x})$

I presume you meant $\frac 3 2 \cos(\frac x 2)$. As inline text, this should be written as (3/2) cos(x/2).

 

[Parsifal1]

I'm moving on to integration now. Here is my attempt at the first question:

f(x)=sin 3x dx

y=-cos 3x^2/2

 

[fresh_42]

I'm moving on to integration now. Here is my attempt at the first question:

f(x)=sin 3x dx

y=-cos 3x^2/2

Control the equation Check your answer by differentiation. Where does the square come from?

 

[Parsifal1]

Control the equation by differentiation. Where does the square come from?

I increased the power of 3x by 1.

 

[fresh_42]

I increased the power of 3x by 1.

Yes, I saw, but why? You want to integrate the sine function, not x.

 

[Parsifal1]

Yes, I saw, but why?

Isn't that hoe you integrate a function? For example 2x---> 2x^2/2

 

[fresh_42]

Isn't that hoe you integrate a function? For example 2x---> 2x^2/2

Yes. Here you have a linear function $x → 2x$ to integrate. But the function you want to integrate is $x → \sin 3x$. You are right that it will give you something with $- \cos$. But what would the derivation of $- \cos \frac{3}{2}x^2$ be?

 

[Parsifal1]

Yes. Here you have a linear function $x → 2x$ to integrate. But the function you want to integrate is $x → \sin 3x$. You are right that it will give you something with $- \cos$. But what would the derivation of $- \cos \frac{3}{2}x^2$ be?

3/2sin^-1*3/2x^2

 

[SteamKing]

3/2sin^-1*3/2x^2

That's not right. It's not even wrong.

 

[Parsifal1]

That's not right. It's not even wrong.

I have no idea how to do it, I know sin goes to -cos, and I thought 3x went to 3x^2/2, that's as far as I got using the maths is fun website.

 

[fresh_42]

3/2sin^-1*3/2x^2

Almost. The chain rule goes: $(f \circ g)'(x) = (f(g(x))' = f'(y) \cdot g'(x)$ where $y= g(x)$ and $f'$ the differentiation according $y$.
So $(- \cos \frac{3}{2} x^2)' = (- \cos y)' \cdot (\frac{3}{2} x^2)' = \sin y \cdot (\frac{3}{2} \cdot 2 \cdot x^{2-1}) = 3x \sin(\frac{3}{2}x^2)$.
The argument in the trigonometric function stays as it is.

 

[Parsifal1]

Almost. The chain rule goes: $(f \circ g)'(x) = (f(g(x))' = f'(y) \cdot g'(x)$ where $y= g(x)$ and $f'$ the differentiation according $y$.
So $(- \cos \frac{3}{2} x^2)' = (- \cos y)' \cdot (\frac{3}{2} x^2)' = \sin y \cdot (\frac{3}{2} \cdot 2 \cdot x^{2-1}) = 3x \sin(\frac{3}{2}x^2)$.
The argument in the trigonometric function stays as it is.

Um, could you explain how I'd integrate that first question a bit more simply, that explanation went right over my head.

 

[SteamKing]

Earlier, you had posted:

I'm revising differentiation for tomorrow, I am hopelessly stuck on some things.

I have some examples with answers, but I have no idea which rules are being used.

y=2sin3x

dy/dx=6cos3x

Now, if you wanted to integrate dy = 6 cos (3x) dx to recover y, then obviously you don't integrate the argument of the cosine (which is 3x) by itself.

The rule for integrating cosine is

$$\int cos (u) du = sin (u) + C$$

Here u = 3x, which means that du = 3 dx. If we plug these quantities back into the integral formula, then we get:

$$\int cos(3x) ⋅ 3\,dx$$

but we are looking to integrate $\int 6 ⋅ cos (3x)\,dx$. We can rewrite the integral above as:

$$2\int cos(3x) ⋅ 3\,dx = 2 sin (3x) + C$$
which is what we were looking to recover, setting C = 0 of course.

 

[SammyS]

I'm moving on to integration now. Here is my attempt at the first question:

f(x)=sin 3x dx

y=-cos 3x^2/2

At this point, you should have started a new thread.

 

[fresh_42]

Um, could you explain how I'd integrate that first question a bit more simply, that explanation went right over my head.

Perhaps I should have written it with $d$'s.
$$\frac{d}{dx} f(g(x)) = \frac{d}{dy} f(y) \cdot \frac{d}{dx} g(x)$$
When you differentiate an antiderivative $F = \int fdx + const$ of a function $f$ you should get $f$ again. So you can always do the test: $\frac{d}{dx}F(x) = f(x)$.

If you differentiate $\sin(3x)$ then you get (according to the chain rule) $\cos(3x) \cdot 3$. The first factor is the derivative of $\sin$ and the second of $3x$.

Now you want to integrate $f(x) = \sin(3x)$, I assume.
Imagine you have integrated the sine function $\sin (y)$, so you get as you said $- \cos (y)$.
Here $y = 3x$. So $\int sin(3x) \, dx ≈ (- \cos (3x))$. But the differentiation for control gives $\frac{d}{dx} (- \cos(3x)) = 3 \cdot \sin(3x)$ according to the chain rule. So there is a correction of a factor $3$ to be made on one side of the $≈$ to make it equal.

 

[Mark44]

@Parsifal1, you should not move on to integration until you o get a better understanding of how the chain rule works in differentiation. Without this understanding, you won't be able to correctly integrate expressions such as sin(3x).

When you integrate, you should always check your answer by differentiating it. If you get back to the original integrand, your work is probably OK.