How do you find the derivative of #arccos(sinx^3)#?

1 Answer
Oct 10, 2016

Use the chain rule, twice

#(d[cos^-1(sin(x^3))])/dx = -3x^2#

Explanation:

The chain rule is:

#(d[f(g(x))])/dx = ((df(g))/(dg))((dg(x))/dx)#

Let #g(x) = sin(x^3)#, then #f(g) = cos^-1(g)# and #(df(g))/(dg) = -1/sqrt(1 - g^2#

For #(dg(x))/dx#, use the chain rule, again:

#(d[g(h(x))])/dx = ((dg(h))/(dh))((dh(x))/dx)#

Let #h(x) = x^3#, then #g(h) = sin(h)#, #(dg(h))/(dh) = cos(h)#, and #(dh(x))/dx = 3x^2#

Substitute back into the second chain rule:

#(dg(x))/dx = 3x^2cos(h)#

Reverse the substitution for h:

#(dg(x))/dx = 3x^2cos(x^3)#

Substitute back into the first chain rule:

#(d[f(g(x))])/dx = -3x^2cos(x^3)/sqrt(1 - g^2)#

Reverse the substitution for g:

#(d[cos^-1(sin(x^3))])/dx = -3x^2cos(x^3)/sqrt(1 - sin^2(x^3))#

Please observe that #sqrt(1 - sin^2(x^3)) = cos(x^3)# which causes the fraction to become 1:

#(d[cos^-1(sin(x^3))])/dx = -3x^2#