Math / Calculus

QuestionFind $f^{\prime}$ , given $f^{\prime}(x)=1, f(x)=\frac{x \sin ^{3}\left(x^{2}\right)}{\sqrt{x^{5}+1}}$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function that's a fraction with $x$ , a sine function, and a square root. Watch out! Don't forget the chain rule and the product rule!
Also, remember to simplify your answer as much as possible.

STEP 2

1. Rewrite the function
2. Calculate the derivative

STEP 3

Let's rewrite the function $f(x)$ to make it easier to differentiate.
We can rewrite the square root as a power of $\frac{1}{2}$ : $f(x) = \frac{x \sin^3(x^2)}{(x^5 + 1)^{\frac{1}{2}}} = x \cdot \sin^3(x^2) \cdot (x^5 + 1)^{-\frac{1}{2}}$ This form is much easier to work with when applying the product and chain rules!

STEP 4

To make things even clearer, let's define: $a(x) = x$ $b(x) = \sin^3(x^2)$ $c(x) = (x^5 + 1)^{-\frac{1}{2}}$ So, $f(x) = a(x) \cdot b(x) \cdot c(x)$ .

STEP 5

Now, let's find the derivatives of $a(x)$ , $b(x)$ , and $c(x)$ : $a'(x) = 1$ For $b'(x)$ , we use the chain rule: $b'(x) = 3\sin^2(x^2) \cdot \cos(x^2) \cdot 2x = 6x\sin^2(x^2)\cos(x^2)$ And for $c'(x)$ , we also use the chain rule: $c'(x) = -\frac{1}{2}(x^5 + 1)^{-\frac{3}{2}} \cdot 5x^4 = -\frac{5x^4}{2(x^5 + 1)^{\frac{3}{2}}}$

STEP 6

Using the product rule for three functions, the derivative of $f(x)$ is: $f'(x) = a'(x)b(x)c(x) + a(x)b'(x)c(x) + a(x)b(x)c'(x)$ Substituting the functions and their derivatives: $f'(x) = 1 \cdot \sin^3(x^2) \cdot (x^5 + 1)^{-\frac{1}{2}} + x \cdot 6x\sin^2(x^2)\cos(x^2) \cdot (x^5 + 1)^{-\frac{1}{2}} + x \cdot \sin^3(x^2) \cdot -\frac{5x^4}{2(x^5 + 1)^{\frac{3}{2}}}$

STEP 7

Let's simplify this expression: $f'(x) = \frac{\sin^3(x^2)}{\sqrt{x^5 + 1}} + \frac{6x^2\sin^2(x^2)\cos(x^2)}{\sqrt{x^5 + 1}} - \frac{5x^5\sin^3(x^2)}{2(x^5 + 1)^{\frac{3}{2}}}$

STEP 8

Our final answer for $f'(x)$ is: $f'(x) = \frac{\sin^3(x^2)}{\sqrt{x^5 + 1}} + \frac{6x^2\sin^2(x^2)\cos(x^2)}{\sqrt{x^5 + 1}} - \frac{5x^5\sin^3(x^2)}{2(x^5 + 1)^{\frac{3}{2}}}$

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