# Differentiation and Integration?

#### TJS1996

##### New member
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

Staff member
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

##### Well-known member
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

##### Well-known member
MHB Math Scholar
... 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

Staff member
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

##### Well-known member
MHB Math Helper
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...