Studdy AI Logo
Math  /  Calculus

QuestionExample: Differentiate the following functions.
1. f(x)=(x4+3)50f(x)=\left(x^{4}+3\right)^{50}
2. g(t)=t3sin⁡tg(t)=\sqrt{t^{3} \sin t}

Studdy Solution

STEP 1

What is this asking? We need to find the *derivatives* of two functions, which tells us how they're changing at any given point. Watch out! Don't forget the chain rule, and remember your trig derivatives!

STEP 2

1. Differentiate f(x)f(x)
2. Differentiate g(t)g(t)

STEP 3

Alright, let's **tackle** f(x)=(x4+3)50f(x) = (x^4 + 3)^{50}!
This screams **chain rule**, doesn't it?
We've got an *inner* function, x4+3x^4 + 3, nestled inside an *outer* function, something to the power of 50.

STEP 4

The chain rule says 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.
So, let's **do this**!

STEP 5

The derivative of the outer function (something to the power of 50) is 50⋅(something)4950 \cdot (\text{something})^{49}.
In our case, "something" is x4+3x^4 + 3, so we have 50(x4+3)4950(x^4 + 3)^{49}.

STEP 6

Now, let's **find** the derivative of the inner function, x4+3x^4 + 3.
That's a simple power rule application, giving us 4x34x^3.

STEP 7

**Multiply** the two results together, and we've got our derivative! f′(x)=50(x4+3)49⋅4x3=200x3(x4+3)49f'(x) = 50(x^4 + 3)^{49} \cdot 4x^3 = \mathbf{200x^3(x^4 + 3)^{49}} Look at that, **beautiful**!

STEP 8

Now, for g(t)=t3sin⁡tg(t) = \sqrt{t^3 \sin t}.
Let's **rewrite** this as g(t)=(t3sin⁡t)1/2g(t) = (t^3 \sin t)^{1/2}.
This makes it clearer that we'll need the **chain rule** again!

STEP 9

Our outer function is something to the power of 1/21/2, and the inner function is t3sin⁡tt^3 \sin t.
The derivative of the outer function is 12(something)−1/2\frac{1}{2}(\text{something})^{-1/2}.

STEP 10

Substituting our inner function, we get 12(t3sin⁡t)−1/2\frac{1}{2}(t^3 \sin t)^{-1/2}.

STEP 11

Now, for the derivative of the inner function, t3sin⁡tt^3 \sin t.
Here, we need the **product rule**!
The product rule says the derivative of two functions multiplied together is the first function times the derivative of the second, plus the second function times the derivative of the first.

STEP 12

The derivative of t3t^3 is 3t23t^2, and the derivative of sin⁡t\sin t is cos⁡t\cos t.
Applying the product rule, we get t3cos⁡t+sin⁡t⋅3t2t^3 \cos t + \sin t \cdot 3t^2.

STEP 13

Finally, **multiply** the derivative of the outer function (evaluated at the inner function) by the derivative of the inner function: g′(t)=12(t3sin⁡t)−1/2⋅(t3cos⁡t+3t2sin⁡t)g'(t) = \frac{1}{2}(t^3 \sin t)^{-1/2} \cdot (t^3 \cos t + 3t^2 \sin t) g′(t)=t3cos⁡t+3t2sin⁡t2t3sin⁡tg'(t) = \mathbf{\frac{t^3 \cos t + 3t^2 \sin t}{2\sqrt{t^3 \sin t}}}**Nailed it**!

STEP 14

The derivative of f(x)f(x) is 200x3(x4+3)49200x^3(x^4 + 3)^{49}, and the derivative of g(t)g(t) is t3cos⁡t+3t2sin⁡t2t3sin⁡t\frac{t^3 \cos t + 3t^2 \sin t}{2\sqrt{t^3 \sin t}}.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord