Question

using differentiation from first principles, find the solution to:
y=sin(2x+1)

Answer 1

You mean find the derivative, yes?, not the solution.

Answer 2

and if so, the derivative in general, or at a point?

Answer 3

in general

Answer 4

What is your first principles definition of derivative?

Answer 5

\[dy/dx=\lim_{h \rightarrow 0}[f(x+h)-f(x)]\div(h)\]

Answer 6

Hence for your function this ratio is
\[ \frac{\sin(2(x+h)+1) - \sin(2x+1)}{h} \]
\[ = \frac{2\sin\left(\frac{2(x+h)+1 + (2x+1)}{2}\right).\cos\left(\frac{2(x+h)+1 - (2x+1)}{2}\right)}{h} \]
Now you take it from here.

Answer 7

please complete

Answer 8

\[\int\limits_{}^{}\cos ^{4}xdx\]

Answer 9

sign error. Should be
\[ = \frac{2\sin([2(x+h)+1 - (2x+1)]/2).\cos([2(x+h)+1+(2x+1)]/2)}{h} \]
\[ = 2\frac{\sin(h)}{h} . \cos(2x + 1 + h)\]

Now take the limit of that as h --> 0.

Answer 10

and remember that
\[ \lim_{h \rightarrow 0} \frac{\sin h}{h} = 1 \]

Answer 11

good

Answer 12

;-)

Answer 13

i'll answer a specific qn if i've time and interest

Answer 14

where's your question?