How do you differentiate #y = (sin(3x) + cot(x^3))^8#?

1 Answer
Jul 31, 2015

Answer:

#y^' = 24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc(x^3)]#

Explanation:

Notice that you can write your function #y# as

#y = u^8#, with #u =sin(3x) + cot(x^3)#

This means that you can use the power rule and the chain rule to differentiate #y#.

For a function #y# that depends on a variable #u#, which in turn depends on a variable #x#, you can determine its derivative by using the chain rule

#color(blue)(d/dx(y) = d/(du)(y) * d/dx(u))#

In your case, this would be equivalent to

#y^' = d/(du)(u^8) * d/dx(u)#

#y^' = 8u^7 * d/dx underbrace((sin(3x) + cot(x^3)))_(color(red)(a(x)))#

Now focus on differentiating function #color(red)(a(x))#, which can be written as

#d/dx(a) = d/dx(sin(3x)) + d/dx(cot(x^3))#

Once again, use the chain rule to differentiate these functions. Remember that

#d/dx(sinx) = cosx#

and that

#d/dx(cotx) = -csc^2x#

The first one will be

#d/dx(sinu_1) = d/(du_1) * (sinu_1) * d/dx(u_1)#, where #u_1 = 3x#

#d/dx(sinu_1) = cosu_1 * d/dx(3x)#

#d/dx(sin(3x)) = 3cos(3x)#

The second one will be

#d/dx(cotu_2) = d/(du_2)(cotu_2) * d/dx(u_2)#, where #u_2 = x^3#

#d/dx(cotu_2) = -csc^2u_2 * d/dx(x^3)#

#d/dx(cotu_2) = -csc^2(x^3) * 3x^2#

Plug these back into your tager derivative to get

#y^' = 8 * [sin(3x) + cot(x^3)]^7 * (3cos(3x) -csc^2(x^3) * 3x^2)#

#y^' = color(green)(24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc(x^3)])#