Question

How do you differentiate `y=5x^6-sec^-1x`?

Answer 1

`dy/dx=30x^5-1/(xsqrt(x^2-1)`

Explanation:

`y=5x^6-sec^-1x`
`dy/dx=d/dx(5x^6-sec^-1x)`
Using sum rule
`=d/dx(5x^6)-d/dx(sec^-1x)`
`d/dx(5x^6)=5.d/dx(x^6)`
`=(5)(6)x^(6-1)=30x^5`
`d/dx(sec^-1x)=1/(xsqrt(x^2-1)`
Thus,
`dy/dx=30x^5-1/(xsqrt(x^2-1)`