How do you use the chain rule to differentiate #y=root3(csc^5 7)#?

Question

Emilia Aguilar · Accepted Answer

#0# No need for chain rule. We know that #y=root(3)(csc^5(7)# will differentiate to #0#, as #csc^5(7)# will give out a number, and so the whole thing will become #0# once differentiated.

Adam Adkins · Answer

undefined To differentiate $ y = \sqrt{3}\csc^5(7x) $ using the chain rule, follow these steps:

1. Identify the outer function and the inner function. In this case, the outer function is the square root function ($\sqrt{3}$) and the inner function is $\csc^5(7x)$.

2. Differentiate the outer function with respect to its variable. The derivative of the square root function $ \sqrt{u} $ with respect to $ u $ is $ \frac{1}{2\sqrt{u}} $.

3. Leave the inner function unchanged.

4. Multiply by the derivative of the inner function with respect to its variable. To differentiate $ \csc^5(7x) $, use the chain rule. The derivative of $ \csc(u) $ with respect to $ u $ is $ -\csc(u) \cot(u) $, and then multiply by the derivative of the inner function with respect to its variable ($7$ in this case).

Combining these steps, the derivative of $ y $ with respect to $ x $ is:

$$ \frac{dy}{dx} = \frac{1}{2\sqrt{\csc^5(7x)}} \times (-5\csc^4(7x)\cot(7x) \times 7) $$

$$ = -\frac{35\csc^4(7x)\cot(7x)}{2\sqrt{\csc^5(7x)}} $$