Differentiation of Trigonoometric Functions

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(1) Let u=sin(5x), du/dx=5cos(5x); dy/dx=(dy/du)(du/dx), y=u-⅓u³, dy/dx=(1-u²)(5cos(5x)), dy/dx=5cos(5x)-5sin²(5x)cos(5x)=5cos(5x)(1-sin²(5x))=5cos³(5x).

(2) y=sin²(4x)+½(1-2sin²(4x))=sin²(4x)+½-sin²(4x)=½, dy/dx=0; also:

dy/dx=8sin(4x)cos(4x)-4sin(8x)=4sin(8x)-4sin(8x)=0

(3) Let u=1-cos(4x), v=sin(4x);

dy/dx=d(u/v)/dx=(vdu/dx-udv/dx)/v²=

(sin(4x))(4sin(4x))-(1-cos(4x))(4cos(4x))/(sin²(4x))=

(4sin²(4x)-4cos(4x)+4cos²(4x))/(sin²(4x))=(4-4cos(4x))/(sin²(4x))=4csc²(4x)-4cot(4x)csc(4x))=

4csc(4x)(csc(4x)-cot(4x)). Also:

y=csc(4x)-cot(4x), dy/dx=-4cot(4x)csc(4x)+4csc²(4x)=4csc(4x)(csc(4x)-cot(4x)).

(4) dy/dx=6sec²(2x)+6tan²(2x)sec²(2x)=

6sec²(2x)(1+tan²(2x))=6sec⁴(2x).

(5) dy/dx=(2sec(4x))(4sec(4x)tan(4x))+(2tan(4x))(4sec²(4x))=

8sec²(4x)tan(4x)+8sec²(4x)tan(4x)=16sec²(4x)tan(4x); also:

y=1+tan²(4x)+tan²(4x)=1+2tan²(4x); dy/dx=(4tan(4x))(4sec²(4x))=16sec²(4x)tan(4x).

 

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