How do you use the chain rule to differentiate #y=(5x^3-3)^5root4(-4x^5-3)#?

Question

Luke Alvarez · Accepted Answer

#dy/dx=75x^2(5x^3-3)^4*(-4x^5-3)^(1/4)-(5x^4*(5x^3-3)^5)/(-4x^5-3)^(3/4)#

First, we need to know two rules of differentiation. The first is the chain rule:

#d/dx(f(g(x)))=f'(g(x))*g'(x)#

The second rule we need is the product rule, which says:

#d/dx(f(x)*g(x))=f'(x)*g(x)+f(x)*g'(x)#

My first step would be to rewrite #root(4)(-4x^5-3)# as a power, rather than a radical. This can be easily done as:

#root(4)(-4x^5-3)=(4x^5-3)^(1/4)#

So now we want to use the chain rule to find the derivative of:

#y=(5x^3-3)^5*(-4x^5-3)^(1/4)#

Now, let's find the derivative of each part, as if it were its own function.

#d/dx((5x^3-3)^5)# #=5(5x^3-3)^4*15x^2#

As you can see, the inside function stays the same. We just use the power rule (bring the power forward, take one off the power), then we use the chain rule, which means we're multiplying by the derivative of the inside function. In the next step, I simplify it.

#=75x^2(5x^3-3)^4#

Now, we'l take the derivative of the other part:

#d/dx((-4x^5-3)^(1/4))# =#1/4(-4x^5-3)^(-3/4)*(-20)x^4#

And simplified further, we can get:

#=-5x^4*(-4x^5-3)^(-3/4)#

Now that we have the pieces to the puzzle, it's time to put them together. Remember, the product rule says: for a function written as:

#y=f(x)*g(x)#

The derivative is written as:

#dy/dx=f'(x)g(x)+f(x)g'(x)#

So let's plug in our functions into this to get the answer:

#dy/dx=75x^2(5x^3-3)^4*(-4x^5-3)^(1/4)+(-5)x^4*(-4x^5-3)^(-3/4)*(5x^3-3)^5#

That's your answer! It doesn't look very clean, and that's because we haven't fully simplified this. First, we can see we're adding a #-5(x^4)#, which is the same as subtracting #5x^4#. So we can simplify that.

#dy/dx=75x^2(5x^3-3)^4*(-4x^5-3)^(1/4)-5x^4*(-4x^5-3)^(-3/4)*(5x^3-3)^5#

We also have a piece with a negative exponent. The rule with this is:

#x^-a=1/x^a#

Therefore, we can say:

#dy/dx=75x^2(5x^3-3)^4*(-4x^5-3)^(1/4)-(5x^4*(5x^3-3)^5)/(-4x^5-3)^(3/4)#

This is the furthest simplification I can think of. But, it'll be your answer.

Charles Anctil · Answer

undefined To differentiate $ y = (5x^3 - 3)^{\frac{5}{4}} \cdot (-4x^5 - 3) $ using the chain rule, follow these steps:

1. Let $ u = 5x^3 - 3 $.
2. Let $ v = -4x^5 - 3 $.
3. Compute $ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} + \frac{dy}{dv} \cdot \frac{dv}{dx} $.
4. Calculate $ \frac{dy}{du} $ by differentiating $ u^{\frac{5}{4}} $ with respect to $ u $.
5. Calculate $ \frac{dy}{dv} $ by differentiating $ v $ with respect to $ v $.
6. Substitute the values of $ \frac{dy}{du} $, $ \frac{du}{dx} $, $ \frac{dy}{dv} $, and $ \frac{dv}{dx} $ back into the chain rule formula.
7. Simplify the expression to get the final result for $ \frac{dy}{dx} $.

Would you like a step-by-step breakdown of these calculations?

William Abdalla · Answer

undefined To differentiate $ y = (5x^3 - 3)^{\frac{1}{4}} (-4x^5 - 3) $ using the chain rule, you follow these steps:

1. Identify the outer function and the inner function. In this case, the outer function is the power function $(\cdot)^{\frac{1}{4}}$ and the inner function is $5x^3 - 3$ for the first term, and $-4x^5 - 3$ for the second term.

2. Differentiate the outer function with respect to its inner function, then multiply by the derivative of the inner function.

3. Put these results together using the chain rule.

So, for the first term:  
- Derivative of the outer function: $ \frac{1}{4}(5x^3 - 3)^{\frac{-3}{4}} $
- Derivative of the inner function: $ 15x^2 $

For the second term:  
- Derivative of the outer function: $ 1 $ (since it's just a constant)
- Derivative of the inner function: $ -20x^4 $

Putting it all together:  
$ \frac{dy}{dx} = \frac{1}{4}(5x^3 - 3)^{\frac{-3}{4}} \cdot 15x^2 \cdot (-4x^5 - 3) + (5x^3 - 3)^{\frac{1}{4}} \cdot (-20x^4) $