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

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Answer 1 · 2016-10-10T19:27:19

Use the chain rule, twice

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

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:

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

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