Question

Find dy/dx by implicit differentiation. y cos x = 5x2 + 3y2

Answer 1

Step 1:
Start by putting $\frac{d}{dx}$ in front of each term

$\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]$
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Step 2:

Deal with the terms in 'x' and the constant terms
$\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]$
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Step 3:

Use the chain rule for the terms in 'y'
$\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}$
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Step 4:

Use the product rule on the term in 'x' and 'y'
$(y) \frac{d}{dx} cos x+(cos x) \frac{d}{dx}y =10x+6y \frac{dy}{dx}$
$y(-siny)+(cosx) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
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Step 5:

Rearrange to make $\frac{dy}{dx}$ the subject
$-y sin(y)+cos(x) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
$cos(x) \frac{dy}{dx}-6y \frac{dy}{dx}=10x+y sin(y)$
$[cos(x) - 6y] \frac{dy}{dx}=10x + y sin(y)$
$\frac{dy}{dx}= \frac{10x+ysin(y)}{cos(x)-6y}$ ⇒ Final Answer