How do you differentiate #y = x^2(sinx)^4 + x(cosx)^-2#?

1 Answer
Oct 16, 2015

Answer:

#dy/dx = 2x(sin x)^4 + 4x^2(sin x)^3*cos x + (cos x)^-2-3x(cos x)^-2*sin x#

Explanation:

#dy/dx = d/dx(x^2(sin x)^4 + x(cos x)^-2)# (differentiate both parts)
#<=>dy/dx = d/dx(x^2(sin x)^4) + d/dx(x(cos x)^-2)# (derivative of sum is sum of derivatives)
#<=>dy/dx = (d/dx(x^2)*(sin x)^4 + x^2*d/dx((sin x)^4)) + d/dx(x(cos x)^-2)#(product rule)
#<=>dy/dx = (2x(sin x)^4 + x^2*4*(sin x)^3*d/dx(sin x)) + d/dx(x(cos x)^-2)# (power rule and chain rule)
#<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + d/dx(x(cos x)^-2)# (trigonometric derivative)
#<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + (d/dx(x)*(cos x)^-2 + x*d/dx((cos x)^-3))# (product rule)
#<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + ((cos x)^-2 + -3*x*(cos x)^-2*d/dx(cos x))# (power rule and chain rule)
#<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + ((cos x)^-2 + -3*x*(cos x)^-2*sin(x))# (trigonometric derivative)

So the derivative is:
#2x(sin x)^4 + 4x^2(sin x)^3*cos x + (cos x)^-2-3x(cos x)^-2*sin x#

You can try simplifying this, but I don't think there's much to simplify.