How do you find the derivative of #y=sin^2x+2^sinx#?

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Answer 1 · 2017-08-04T11:43:15

# dy/dx=sin2x+2^sinx*cosx*ln2.#

#y=sin^2x+2^sinx=(sinx)^2+2^sinx.#

#:. dy/dx=d/dx{(sinx)^2}+d/dx{2^sinx}......(1).#

Using the Chain Rule,

#d/dx{(sinx)^2}=2*(sinx)^(2-1)*d/dx{sinx},#

#=2sinxcosx=sin2x............................(2).#

Further, we know that, #d/dx{a^x}=(a^x)lna.#

Reusing the Chain Rule,

#d/dx{2^sinx}=(2^sinx)ln2d/dx{sinx},#

#=2^sinx*cosx*ln2..................................................(3).#

From #(1),(2), and, (3),# we have,

# dy/dx=sin2x+2^sinx*cosx*ln2.#