How do you use the chain rule to differentiate #y=(x^3+3)^5#?

Question

Elijah Abram · Accepted Answer

#dy/dx=5(x^3+3)^4(3x^2)# Let #y=u^n(x)# So #dy/dx=n*u^(n-1)(x)*u'(x)# so applying this #dy/dx=5(x^3+3)^4(3x^2)#

Paisley Johnson · Answer

undefined To differentiate $ y = (x^3 + 3)^5 $ using the chain rule, follow these steps:

1. Identify the outer function: $ u = (x^3 + 3) $
2. Identify the inner function: $ v = x^3 + 3 $
3. Differentiate the inner function with respect to $ x $: $ \frac{dv}{dx} = 3x^2 $
4. Substitute the inner function into the outer function: $ y = u^5 $
5. Apply the chain rule: $ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} $
6. Find $ \frac{dy}{du} $ by differentiating $ u^5 $: $ \frac{dy}{du} = 5u^4 $
7. Substitute $ u = (x^3 + 3) $ and $ \frac{du}{dx} = \frac{dv}{dx} = 3x^2 $ into $ \frac{dy}{dx} $
8. Simplify the expression: $ \frac{dy}{dx} = 5(x^3 + 3)^4 \times 3x^2 $
9. Further simplify: $ \frac{dy}{dx} = 15x^2(x^3 + 3)^4 $

So, $ \frac{dy}{dx} = 15x^2(x^3 + 3)^4 $