Question

Given: x = tan4t and y = sec4t. Find y’.

Answer 1

$$y'_x=\dfrac{y'_t}{x'_t}$$

$$x'_t=4\big(\dfrac{1}{\cos^2(4t)}\big)$$

$$y'_t=4\big(-\dfrac{1}{\cos^2(4t)}\big)(-\sin(4t)$$

$$y'_x=\dfrac{y'_t}{x'_t}=\dfrac{\dfrac{4\sin (4t)}{\cos^2(4t)}}{\dfrac{4}{\cos^2(4t)}}=\sin (4t)$$

$$y'_x=\sin(4t)$$