Differentiate $(a^2sin^2x+b^2cos^2x)$ ?

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Answer 1 · 2016-09-08T15:28:47

$d/(dx) (a^2sin^2x+b^2cos^2x)=sin2x(a^2-b^2)$

$d/(dx) (a^2sin^2x+b^2cos^2x)$

= $d/(dx) a^2sin^2x+d/(dx) b^2cos^2x$

= $a^2xx2sinx xx cosx+b^2xx2cosx xx(-sinx)$

= $2a^2sinxcosx-2b^2sinxcosx$

= $sin2x(a^2-b^2)$

Differentiate (a^2sin^2x+b^2cos^2x)(a^2sin^2x+b^2cos^2x)?