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Math  /  Calculus

QuestionFind f′(x)f^{\prime}(x). f(x)=(5x6+6)3f′(x)=□\begin{array}{l} f(x)=\left(5 x^{6}+6\right)^{3} \\ f^{\prime}(x)=\square \end{array}

Studdy Solution

STEP 1

What is this asking? Find the *derivative* of a function that's a *polynomial* raised to a *power*. Watch out! Don't forget the *chain rule*!

STEP 2

1. Set up the chain rule
2. Compute the outer derivative
3. Compute the inner derivative
4. Combine the results

STEP 3

Alright, so we're given the function f(x)=(5x6+6)3f(x) = (5x^6 + 6)^3.
We want to find its derivative, which we write as f′(x)f'(x).
Since we have a function inside another function (a *composite function*), we'll use the **chain rule**.
Think of it like peeling an onion, layer by layer!

STEP 4

The chain rule says that the derivative of a composite function is the derivative of the *outer* function, evaluated at the *inner* function, *times* the derivative of the *inner* function.
Let's write our function as f(x)=g(h(x))f(x) = g(h(x)), where g(u)=u3g(u) = u^3 is the *outer* function and h(x)=5x6+6h(x) = 5x^6 + 6 is the *inner* function.

STEP 5

First, let's find the derivative of the outer function, g(u)=u3g(u) = u^3.
Using the power rule, we get g′(u)=3u2g'(u) = 3u^2.
Remember, the power rule says that the derivative of unu^n is nun−1nu^{n-1}.

STEP 6

Now, we need to evaluate this derivative at the inner function, h(x)h(x).
So we substitute h(x)h(x) for uu in g′(u)g'(u).
This gives us g′(h(x))=3(5x6+6)2g'(h(x)) = 3(5x^6 + 6)^2.
See how we just plugged in the inner function?

STEP 7

Next, we find the derivative of the inner function, h(x)=5x6+6h(x) = 5x^6 + 6.
The derivative of 5x65x^6 is 30x530x^5 (using the power rule again: 6⋅5x6−16 \cdot 5x^{6-1}), and the derivative of a constant (like 6) is **zero**.
So, h′(x)=30x5+0=30x5h'(x) = 30x^5 + 0 = 30x^5.

STEP 8

Finally, we multiply the outer derivative evaluated at the inner function by the inner derivative: f′(x)=g′(h(x))⋅h′(x)f'(x) = g'(h(x)) \cdot h'(x).

STEP 9

Plugging in our results, we get f′(x)=3(5x6+6)2⋅30x5f'(x) = 3(5x^6 + 6)^2 \cdot 30x^5.

STEP 10

Simplifying, we have f′(x)=90x5(5x6+6)2f'(x) = 90x^5(5x^6 + 6)^2.
And there you have it!

STEP 11

f′(x)=90x5(5x6+6)2f'(x) = 90x^5(5x^6 + 6)^2

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