How do you differentiate given #y = (x^2 tan x) / (sec x)#?

Question

Emma Aldridge · Accepted Answer

#(dy)/(dx)=2xsinx+x^2cosx# #y=(x^2tanx)/secx=(x^2sinx/cosx)/(1/cosx)=x^2sinx# Now using product rule #(dy)/(dx)=2xsinx+x^2cosx#

Ezekiel Alexander · Answer

undefined To differentiate $ y = \frac{x^2 \tan x}{\sec x} $, you can use the quotient rule, which states that if $ y = \frac{u}{v} $, then $ y' = \frac{u'v - uv'}{v^2} $.

Here's the step-by-step process:

1. Let $ u = x^2 \tan x $ and $ v = \sec x $.
2. Find $ u' $ and $ v' $.
   - $ u' = 2x \tan x + x^2 \sec^2 x $ (product rule and derivative of tangent)
   - $ v' = \sec x \tan x $ (derivative of secant)
3. Plug these into the quotient rule formula:
   $$
   y' = \frac{(2x \tan x + x^2 \sec^2 x) \cdot \sec x - (x^2 \tan x) \cdot (\sec x \tan x)}{(\sec x)^2}
   $$
4. Simplify the expression:
   $$
   y' = \frac{2x \sec x \tan^2 x + x^2 \sec^3 x - x^2 \sec x \tan^2 x}{\sec^2 x}
   $$
5. Further simplify by factoring out common terms:
   $$
   y' = \frac{x^2 \sec x (\sec^2 x + 2 \tan^2 x) - 2x \sec x \tan^2 x}{\sec^2 x}
   $$
6. Finally, simplify the expression to get the derivative:
   $$
   y' = x^2 (\sec^2 x + 2 \tan^2 x) - 2x \tan^2 x
   $$

So, the derivative of $ y = \frac{x^2 \tan x}{\sec x} $ is $ y' = x^2 (\sec^2 x + 2 \tan^2 x) - 2x \tan^2 x $.