What is the derivative of #sin(x^2+5) cos(x^2+9x+2)#?

1 Answer
Stefan V.
Sep 9, 2015

#x * cos(x^2 + 5) * cos(x^2 + 9x + 2) - (2x+9) * sin(x^2+5) * sin(x^2 + 9x + 2)#

Explanation:

You can differentiate this function by using the product rule

#color(blue)(d/dx(f(x) * g(x)) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x))#

the fact that

#d/dx(sinx) = cosx" "# and #" "d/dx(cosx) = -sinx#

and the chain rule for #sinu#, with #u = x^2 + 5# and #cost#, with #t = x^2 + 9x + 2#.

So, put all this together to get

#y = sinu * sin t#

#d/dx(y) = [d/dx(sinu)] * cost + sinu * d/dx(cost)#

#y^' = [d/(du)(sinu) * d/dx(u)] * cost + sinu * [d/(dt)(cost) * d/dx(t)]#

#y^' = [cosu * d/dx(x^2 + 5)] * cost + sinu * [(-sint) * d/dx(x^2+9x+2)]#

#y^' = [cos(x^2+5) * 2x] * cos(x^2 + 9x + 2) + sin(x^2 + 5) * [-sin(x^2 + 9x + 2) * (2x + 9)]#

#y^' = color(green)(2x * cos(x^2 + 5) * cos(x^2 + 9x + 2) - (2x+9) * sin(x^2+5) * sin(x^2 + 9x + 2))#

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