Question

help me........trigonometric differentiation

Answer 1

\[y=3/8x+3/8\sin x \cos x+1/4\cos^3sinx\]

Answer 2

help @dpasingh @druminjosh @Gerald_R9 @ybarrap @khurramshahzad @ash2326 @Luigi0210

Answer 3

Is your question like
\[y=\frac{3}{8}x+\frac{3}{8} sinxcosx+\frac{1}{4} \cos3sinx\]

Answer 4

yes...derive that one pls...

Answer 5

\[y=\frac{3}{8}x+\frac{3}{8} sinxcosx+\frac{1}{4} \cos^3sinx\]
Differentiating w.r.t x we find
\[y'=\frac{3}{8}+\frac{3}{8} [sinx \frac{d}{dx}cosx+cosx \frac{d}{dx}sinx]+\frac{1}{4} [\cos^3x \frac{d}{dx}sinx+\sin x \frac{d}{dx}\cos^3 x]\]

Answer 6

\[y=3/8x+3/8sinxcosx+1/4 \cos^3x sinx\]

Answer 7

am i need a more simplified answer...pls,.

Answer 8

\[\[y'=\frac{3}{8}+\frac{3}{8} [sinx (-sinx)+cosx cosx]+\frac{1}{4} [\cos^3x cosx+\sin x 3\cos^2 x (-sinx)]\]
\[y'=\frac{3}{8}+\frac{3}{8} [-\sin^2x +\cos^2x ]+\frac{1}{4} [\cos^4x -\sin ^2x 3\cos^2 x ]\]
\[y'=\frac{3}{8}+\frac{3}{8} [\cos^2x-\sin^2x]+\frac{1}{4} [\cos^4x -3\sin ^2x \cos^2 x ]\]
\[y'=\frac{3}{8}+\frac{3}{8} [\cos2x]+\frac{1}{4} [\cos^4x -3\sin ^2x \cos^2 x ]\]

Answer 9

the final answer must be \[\cos^4x\]

Answer 10

i just want to see the full process of solving

Answer 11

well use some simple trig identities and substitute

say, yo may need to change the subject

\[\sin^2(x) + \cos^2(x) = 1\]

and the other obvious one is

\[\cos(2x) = \cos^2(x) - \sin^2(x).... or \cos(2x) = 2\cos^2(x) - 1\]

then just work on simplifying

Answer 12

am..its not just that easy as you think .........
it is

Answer 13

@campbell_st

Answer 14

well look at it and you'll see that it may well be... making the correct substitution for cos(2x) will see the constant 3/8 disappear

so I'd say yes it is that easy now you have the derivative... you just need to work through the substitutions....

Answer 15

@dpasingh has done a great job with the derivative.... so take advantage of it...

Answer 16

Refer to the attached calculation.

Answer 17

$$y=3/8x+3/8\sin x\cos x+1/4 \cos^3x\sin x\\y'=3/8+3/8\cos^2x-3/8\sin^2 x+1/4\cos^4 x-3/4\cos^2 x\sin^2 x$$recall: \(\cos^2 x-\sin^2 x=\cos2x\) but also \(2\cos x\sin x=\sin2x\) hence:$$y'=3/8+3/8\cos 2x+1/4\cos^4 x-3/16\sin^22x$$next, recall \(1-\sin^2 x=\cos^2 x\) hence:$$y'=3/16+3/8\cos 2x+1/4\cos^4x+3/16\cos^22x\\y'=-3/16+3/8(1+\cos 2x)+1/4\cos^4x+3/16\cos^22x$$next observe we have \(1+\cos2x=2\cos^2x\):$$y'=-3/16+3/4\cos^2x+1/4\cos^4 x+3/16\cos^2 2x\\y'=-3/16(1-\cos^22x)+3/4\cos^2 x+1/4\cos^4x$$next use the fact that \(1-\cos^2x=\sin^2x\):$$y'=-3/16\sin^22x+3/4\cos^2x+1/4\cos^4x$$recall \(\sin2x=2\cos x\sin x\):$$y'=-3/4\cos^2 x\sin^2 x+3/4\cos^2 x+1/4\cos^4 x\\y'=3/4\cos^2 x(1-\sin^2x)+1/4\cos^4x$$again using \(1-\sin^2x=\cos^2x\) we end up with:$$y'=3/4\cos^4x+1/4\cos^4x\\y'=\cos^4x$$