Sure, let's differentiate each function step-by-step.
### Part (i): Differentiate [tex]\( y = x^2 \sin(5x) \)[/tex]
To differentiate [tex]\( y = x^2 \sin(5x) \)[/tex], we'll use the product rule, which states that if you have a function of the form [tex]\( f(x) = u(x)v(x) \)[/tex], then its derivative is [tex]\( f'(x) = u'(x)v(x) + u(x)v'(x) \)[/tex].
Here, let:
- [tex]\( u(x) = x^2 \)[/tex]
- [tex]\( v(x) = \sin(5x) \)[/tex]
First, we find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
- [tex]\( u'(x) = \frac{d}{dx}(x^2) = 2x \)[/tex]
- [tex]\( v'(x) = \frac{d}{dx}(\sin(5x)) = 5\cos(5x) \)[/tex] (using the chain rule)
Now, apply the product rule:
[tex]\[ \frac{d}{dx}(x^2 \sin(5x)) = u'(x)v(x) + u(x)v'(x) \][/tex]
[tex]\[ \frac{d}{dx}(x^2 \sin(5x)) = 2x \sin(5x) + x^2 (5 \cos(5x)) \][/tex]
[tex]\[ \frac{d}{dx}(x^2 \sin(5x)) = 2x \sin(5x) + 5x^2 \cos(5x) \][/tex]
So, the derivative is:
[tex]\[ \frac{d}{dx}(x^2 \sin(5x)) = 5x^2 \cos(5x) + 2x \sin(5x) \][/tex]
### Part (ii): Differentiate [tex]\( y = \frac{2x + 5}{\cos(3x)} \)[/tex]
To differentiate [tex]\( y = \frac{2x + 5}{\cos(3x)} \)[/tex], we'll use the quotient rule, which states that if you have a function of the form [tex]\( f(x) = \frac{u(x)}{v(x)} \)[/tex], then its derivative is:
[tex]\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \][/tex]
Here, let:
- [tex]\( u(x) = 2x + 5 \)[/tex]
- [tex]\( v(x) = \cos(3x) \)[/tex]
First, we find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
- [tex]\( u'(x) = \frac{d}{dx}(2x + 5) = 2 \)[/tex]
- [tex]\( v'(x) = \frac{d}{dx}(\cos(3x)) = -3\sin(3x) \)[/tex] (using the chain rule)
Now, apply the quotient rule:
[tex]\[ \frac{d}{dx}\left(\frac{2x + 5}{\cos(3x)}\right) = \frac{(2)(\cos(3x)) - (2x + 5)(-3\sin(3x))}{(\cos(3x))^2} \][/tex]
[tex]\[ \frac{d}{dx}\left(\frac{2x + 5}{\cos(3x)}\right) = \frac{2\cos(3x) + 3(2x + 5)\sin(3x)}{\cos(3x)^2} \][/tex]
Simplify the numerator:
[tex]\[ 2 \cos(3x) + 3 (2x + 5) \sin(3x) \][/tex]
Therefore, the derivative is:
[tex]\[ \frac{d}{dx}\left(\frac{2x + 5}{\cos(3x)}\right) = \frac{3 (2x + 5) \sin(3x) + 2 \cos(3x)}{\cos(3x)^2} \][/tex]
[tex]\[ \frac{d}{dx}\left(\frac{2x + 5}{\cos(3x)}\right) = \frac{ 3(2x + 5) \sin(3x)}{\cos(3x)^2} + \frac{2}{\cos(3x)} \][/tex]
So, the derivative is:
[tex]\[ \frac{d}{dx}\left(\frac{2x + 5}{\cos(3x)}\right) = 3 \frac{(2x + 5) \sin(3x)}{\cos^2(3x)} + \frac{2}{\cos(3x)} \][/tex]
In summary, the derivatives are:
(i) [tex]\( \frac{d}{dx}(x^2 \sin(5x)) = 5x^2 \cos(5x) + 2x \sin(5x) \)[/tex]
(ii) [tex]\( \frac{d}{dx}\left(\frac{2x+5}{\cos(3x)}\right) = 3 \frac{(2x + 5) \sin(3x)}{\cos^2(3x)} + \frac{2}{\cos(3x)} \)[/tex]
