Math / Calculus

QuestionPart IV Find the $d y$ . ( 5 scores per question. The total is 10 scores.)
15. $y=\sin (\tan x+\cos x)$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the function $y = \sin(\tan x + \cos x)$ .
2. We need to find the differential $dy$ .
3. We will use the chain rule and the derivatives of trigonometric functions to find $dy$ .

STEP 2

To find $dy$ , we first need to find the derivative of $y$ with respect to $x$ , denoted as $\frac{dy}{dx}$ .

STEP 3

Apply the chain rule to differentiate $y = \sin(\tan x + \cos x)$ . The chain rule states that if $y = f(g(x))$ , then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$ .

STEP 4

Identify the outer function $f(u) = \sin(u)$ and the inner function $u = \tan x + \cos x$ .

STEP 5

Differentiate the outer function $f(u) = \sin(u)$ with respect to $u$ . The derivative is $f'(u) = \cos(u)$ .

STEP 6

Differentiate the inner function $u = \tan x + \cos x$ with respect to $x$ .

STEP 7

The derivative of $\tan x$ with respect to $x$ is $\sec^2 x$ .

STEP 8

The derivative of $\cos x$ with respect to $x$ is $-\sin x$ .

STEP 9

Combine the derivatives from STEP_7 and STEP_8 to find $\frac{du}{dx}$ .
$\frac{du}{dx} = \sec^2 x - \sin x$

STEP 10

Apply the chain rule to find $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \cos(\tan x + \cos x) \cdot (\sec^2 x - \sin x)$

STEP 11

Now that we have $\frac{dy}{dx}$ , we can express $dy$ as:
$dy = \frac{dy}{dx} \cdot dx$

STEP 12

Substitute the expression for $\frac{dy}{dx}$ from STEP_10 into the equation for $dy$ .
$dy = \cos(\tan x + \cos x) \cdot (\sec^2 x - \sin x) \cdot dx$
This is the expression for $dy$ .

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