What is #f_x# when #f (x,y) = sin^2 (x^2y^2)# ?

Question

HIX Tutor · Accepted Answer

undefined To find $ f_x $ when $ f(x,y) = \sin^2(x^2y^2) $, follow these steps:

1. **Understand what $ f_x $ means**: It means we want to find the derivative of $ f $ with respect to $ x $.

2. **Use the chain rule**: Since $ f $ is a composition of functions, we can use the chain rule.

3. **Differentiate $ f $**:
   - The outer function is $ \sin^2(u) $, where $ u = x^2y^2 $.
   - The derivative of $ \sin^2(u) $ is $ 2\sin(u)\cos(u) $ (using the chain rule).
   - Now, we also need to find the derivative of $ u = x^2y^2 $ with respect to $ x $. This gives $ 2xy^2 $.

4. **Put it all together**:
   - So, $ f_x = 2\sin(x^2y^2)\cos(x^2y^2) \cdot 2xy^2 $.

5. **Final answer**: Combine it:
   $$
   f_x = 4xy^2 \sin(x^2y^2) \cos(x^2y^2)
   $$

And that's your answer!

William Abdalla · Answer

# rArr f_x=(delf)/(delx)=2xy^2sin(2x^2y^2)#.

#f(x,y)=sin^2(x^2y^2)=[sin{(xy)^2}]^2#. Recall that, # f_x," denoted also as "(delf)/(delx)",# is the derivative of #f# w.r.t. #x,# treating #y,# as constant. Accordingly, #f_x=2sin{(xy)^2}*del/(delx)[sin{(xy)^2}]#, #=2sin(x^2y^2)*cos{(xy)^2}*del/(delx)(xy)^2#, #=color(red)(2sin(x^2y^2)cos(x^2y^2))*del/(delx)(x^2y^2)#, #=color(red)(sin(2x^2y^2))*{y^2del/(delx)(x^2)}....[because, y" is constant]"#, #=sin(2x^2y^2){y^2*2x}#. # rArr f_x=(delf)/(delx)=2xy^2sin(2x^2y^2)#. Enjoy Maths.!

Elijah Abram · Answer

undefined To find $ f_x $ when $ f(x, y) = \sin^2(x^2y^2) $, we need to differentiate $ f $ with respect to $ x $ while treating $ y $ as a constant.

Using the chain rule and the power rule, we first differentiate the outer function $ \sin^2(u) $ with respect to its inner function $ u = x^2y^2 $, and then multiply by the derivative of $ u $ with respect to $ x $.

So, we get:
$$ f_x = 2\sin(x^2y^2)\cos(x^2y^2)(2xy^2) = 4xy^2\sin(x^2y^2)\cos(x^2y^2) $$