Question

How do you determine `(dy)/(dx)` given `y=sec(sqrt(x^3+x))`?

Answer 1

`(dy)/(dx)=sec(sqrt(x^3+x))tan(sqrt(x^3+x))(3x^2+1)/(2sqrt(x^3+x))`

Explanation:

We use chain rule here to differentiate. Let `y=sec(g(x))`, where `g(x)=sqrt(h(x))` and `h(x)=x^3+x`

As such `(dy)/(dg(x))=sec(g(x))tan(g(x))`, `(dg(x))/(dh(x))=1/(2sqrt(h(x))` and `(dh(x))/(dx)=3x^2+1`

Hence `(dy)/(dx)=(dy)/(dg(x)) * (dg(x))/(dh(x))*(dh(x))/(dx)`

= `sec(g(x))tan(g(x)) * 1/(2sqrt(h(x))) * (3x^2+1)`

= `sec(sqrt(h(x)))tan(sqrt(h(x))) * 1/(2sqrt(x^3+x)) * (3x^2+1)`

= `sec(sqrt(x^3+x))tan(sqrt(x^3+x)) * 1/(2sqrt(x^3+x)) * (3x^2+1)`

= `sec(sqrt(x^3+x))tan(sqrt(x^3+x))(3x^2+1)/(2sqrt(x^3+x))`