How do you differentiate  y=sec (3 - 8x) using the chain rule?

Dec 6, 2015

The derivative of secant is $\sec x \tan x$ and the derivative of $3 - 8 x \text{ is} - 8$ ...

Explanation:

$y ' = \sec \left(3 - 8 x\right) \times \tan \left(3 - 8 x\right) \times - 8$

hope that helped

Dec 6, 2015

$y ' = 8 \sec \left(8 x - 3\right) \tan \left(8 x - 3\right)$

Explanation:

According to the chain rule, $\frac{d}{\mathrm{dx}} \left[\sec \left(u\right)\right] = u ' \sec \left(u\right) \tan \left(u\right)$.

Therefore, $\frac{d}{\mathrm{dx}} \left[\sec \left(3 - 8 x\right)\right] = \frac{d}{\mathrm{dx}} \left[3 - 8 x\right] \sec \left(3 - 8 x\right) \tan \left(3 - 8 x\right)$.

$\frac{d}{\mathrm{dx}} \left[3 - 8 x\right] = - 8$, so:

$y ' = - 8 \sec \left(3 - 8 x\right) \tan \left(3 - 8 x\right)$

Another slightly different approach would be to recognize that $\cos \left(- a\right) = \cos \left(a\right)$, and since $\sec \left(a\right) = \frac{1}{\cos} \left(a\right) , \sec \left(- a\right) = \sec \left(a\right)$.

Thus, $\sec \left(3 - 8 x\right) = \sec \left(8 x - 3\right)$, and $y ' = 8 \sec \left(8 x - 3\right) \tan \left(8 x - 3\right)$.