Differentiation and Integration?

[TJS1996]

Really struggling on these 2 questions for a Maths assignment, I've got to find dy/dx. Could anyone help me with the working out and answers please?

a) y=3sin(4x)-5+5cos(x/3)+4x

b) dy/dx=4cos(2x)+6 (given that y(0)=7)

 

[MarkFL]

I have moved this topic to the Calculus sub-forum, as it involves differentiation and integration.

Let's begin with the first problem. I am assuming you know the following rules:

a) \(\displaystyle \frac{d}{dx}(\sin(u(x)))=\cos(u)\frac{du}{dx}\)

b) \(\displaystyle \frac{d}{dx}(\cos(u(x)))=-\sin(u)\frac{du}{dx}\)

c) \(\displaystyle \frac{d}{dx}(kx^n)=knx^{n-1}\)

Can you apply these appropriately to each term?

 

[chisigma]

Really struggling on these 2 questions for a Maths assignment, I've got to find dy/dx. Could anyone help me with the working out and answers please?

b) dy/dx=4cos(2x)+6 (given that y(0)=7)

If You have a differential equation written in the form...

$$f(y)\ dy = g(x)\ dx\ (1)$$

... F(*) is any primitive of f(*) and G(*) any primitive of g(*), then the solution is in the form...

$$F(y) = G(x) + c\ (2)$$

... where c is an arbitrary constant. Can Your equation be written in the form (1) and how?...

Kind regards

$\chi$ $\sigma$

 

[Evgeny.Makarov]

... F(*) is any primitive of f(*) and G(*) any primitive of g(*)

If $F'=f$, then $F$ is usually called an antiderivative, or indefinite integral, of $f$ in English. However, Wikipedia gives "primitive integral" as another possible synonym.

 

[MarkFL]

Really struggling on these 2 questions for a Maths assignment, I've got to find dy/dx. Could anyone help me with the working out and answers please?

a) y=3sin(4x)-5+5cos(x/3)+4x

b) dy/dx=4cos(2x)+6 (given that y(0)=7)

Several months has gone by, and so I will now provide the solutions:

a) Differentiate the following with respect to $x$:

\(\displaystyle y=3\sin(4x)-5+5\cos\left(\frac{x}{3} \right)+4x\)

Differentiating term by term, and applying the chain rule as necessary, we find:

\(\displaystyle \frac{dy}{dx}=3\cos(4x)\cdot4-0-5\sin\left(\frac{x}{3} \right)\cdot\frac{1}{3}+4\)

\(\displaystyle \frac{dy}{dx}=12\cos(4x)-\frac{5}{3}\sin\left(\frac{x}{3} \right)+4\)

b) Solve the following IVP:

\(\displaystyle \frac{dy}{dx}=4\cos(2x)+6\) where \(\displaystyle y(0)=7\)

Switching dummy variables of integration, and using the initial values as the limits of integration, we may write:

\(\displaystyle \int_{y(0)}^{y(x)}\,du=2\int_0^x2\cos(2v)+3\,dv\)

\(\displaystyle y(x)-y(0)=2\left[\sin(2v)+3v \right]_0^x=2\left(\sin(2x)+3x \right)\)

And so the solution satisfying the given conditions is:

\(\displaystyle y(x)=2\left(\sin(2x)+3x \right)+7\)

 

[Prove It]

Really struggling on these 2 questions for a Maths assignment, I've got to find dy/dx. Could anyone help me with the working out and answers please?

a) y=3sin(4x)-5+5cos(x/3)+4x

b) dy/dx=4cos(2x)+6 (given that y(0)=7)

There it is...