Respuesta :
[tex]\displaystyle f(x)=\sin(3x)\\\\\\f'(x)=\lim_{h\to0}\dfrac{\sin(3(x+h))-\sin(3x)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{\sin(3x+3h)-\sin(3x)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{3x+3h+3x}{2}\right)\sin\left(\dfrac{3x+3h-3x}{2}\right)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{6x+3h}{2}\right)\sin\left(\dfrac{3h}{2}\right)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{6x+3h}{2}\right)\sin\left(\dfrac{3h}{2}\right)}{\dfrac{3h}{2}}\cdot\dfrac{3}{2}[/tex]
[tex]\displaystyle f'(x)=\lim_{h\to0}2\cos\left(\dfrac{6x+3h}{2}\right)\cdot\dfrac{3}{2}\\\\f'(x)=\lim_{h\to0}3\cos\left(\dfrac{6x+3h}{2}\right)\\\\f'(x)=3\cos\left(\dfrac{6x+3\cdot 0}{2}\right)\\\\f'(x)=3\cos\left(\dfrac{6x}{2}\right)\\\\f'(x)=3\cos(3x)[/tex]
The derivative of sin 3x using first principles is; 3cos(3x)
We want to find the derivative of sin 3x using first principles.
Step 1;
f'(x) = [tex]\lim_{h \to \ 0} \frac{(sin3(x + h)) - sin(3x)}{h}[/tex]
Step 2; Expand the bracket to get;
f'(x) = [tex]\lim_{h \to \ 0} \frac{(sin(3x + 3h)) - sin(3x)}{h}[/tex]
Step 3: According to trigonometric identities, we know that;
sin A - sin B = [tex]2cos\frac{A + B}{2} sin\frac{A - B}{2}[/tex]
Applying that to the answer in step 2 gives;
f'(x) = [tex]\lim_{h \to \ 0} \frac{2(cos\frac{(3x + 3h + 3x)}{2}) sin\frac{(3x + 3h - 3x)}{2})}{h}[/tex]
Step 4; Simplify the brackets to obtain;
f'(x) = [tex]\lim_{h \to \ 0} \frac{2(cos\frac{(6x + 3h)}{2}) sin\frac{(3h)}{2})}{h}[/tex]
Step 5; Rewrite the denominator to get;
f'(x) = [tex]\lim_{h \to \ 0} \frac{2(cos\frac{(6x + 3h)}{2}) sin\frac{(3h)}{2})}{\frac{3h}{2}*\frac{2}{3}}[/tex]
Step 6; Input the limit of 0 for h to get;
f'(x) = [tex]\frac{2(cos\frac{(6x)}{2}) sin\frac{(0)}{2})}{\frac{0}{2}*\frac{2}{3}}[/tex]
⇒ f'(x) = [tex]\frac{2(cos3x)}{2/3} \frac{ 0}{{0}}[/tex]
0/0 = 1. Thus;
f'(x) = 3cos(3x)
Read more about differentiation from first principles at; https://brainly.com/question/5313449
