Math / Calculus

QuestionScore Part IV Find the $d y$ . ( 5 scores per question. The total is 10 scores.)
15. $y=\sin (\tan x+\cos x)$ .
16. $e^{y}-x^{2} y=0$ .

Studdy Solution

STEP 1

Assumptions
1. We need to find the differential $dy$ for each given function.
2. For problem 15, the function is $y = \sin(\tan x + \cos x)$ .
3. For problem 16, the function is implicitly defined by the equation $e^y - x^2 y = 0$ .
4. We will use differentiation techniques, such as the chain rule and implicit differentiation, to find $dy$ .

STEP 2

For problem 15, we will differentiate $y = \sin(\tan x + \cos x)$ with respect to $x$ .

STEP 3

Apply the chain rule to differentiate $y = \sin(\tan x + \cos x)$ .
$\frac{dy}{dx} = \cos(\tan x + \cos x) \cdot \frac{d}{dx}(\tan x + \cos x)$

STEP 4

Differentiate the inner function $\tan x + \cos x$ with respect to $x$ .
$\frac{d}{dx}(\tan x + \cos x) = \sec^2 x - \sin x$

STEP 5

Substitute the derivative of the inner function back into the chain rule expression.
$\frac{dy}{dx} = \cos(\tan x + \cos x) \cdot (\sec^2 x - \sin x)$

STEP 6

Express $dy$ in terms of $dx$ .
$dy = \cos(\tan x + \cos x) \cdot (\sec^2 x - \sin x) \, dx$

STEP 7

For problem 16, we will use implicit differentiation on the equation $e^y - x^2 y = 0$ .

STEP 8

Differentiate both sides of the equation $e^y - x^2 y = 0$ with respect to $x$ .
$\frac{d}{dx}(e^y) - \frac{d}{dx}(x^2 y) = 0$

STEP 9

Apply the chain rule to differentiate $e^y$ .
$\frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx}$

STEP 10

Apply the product rule to differentiate $x^2 y$ .
$\frac{d}{dx}(x^2 y) = x^2 \cdot \frac{dy}{dx} + 2x \cdot y$

STEP 11

Substitute the derivatives back into the differentiated equation.
$e^y \cdot \frac{dy}{dx} - (x^2 \cdot \frac{dy}{dx} + 2x \cdot y) = 0$

STEP 12

Rearrange the equation to solve for $\frac{dy}{dx}$ .
$e^y \cdot \frac{dy}{dx} - x^2 \cdot \frac{dy}{dx} = 2x \cdot y$

STEP 13

Factor out $\frac{dy}{dx}$ from the left side of the equation.
$(e^y - x^2) \cdot \frac{dy}{dx} = 2x \cdot y$

STEP 14

Solve for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{2x \cdot y}{e^y - x^2}$

STEP 15

Express $dy$ in terms of $dx$ .
$dy = \frac{2x \cdot y}{e^y - x^2} \, dx$
Solution for problem 15: $dy = \cos(\tan x + \cos x) \cdot (\sec^2 x - \sin x) \, dx$
Solution for problem 16: $dy = \frac{2x \cdot y}{e^y - x^2} \, dx$

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