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Fudgin [204]
3 years ago
9

Find dy/dx by implicit differentiation for ysin(y) = xcos(x)

Mathematics
1 answer:
tatyana61 [14]3 years ago
5 0

Answer:

\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}

Step-by-step explanation:

So we have:

y\sin(y)=x\cos(x)

And we want to find dy/dx.

So, let's take the derivative of both sides with respect to x:

\frac{d}{dx}[y\sin(y)]=\frac{d}{dx}[x\cos(x)]

Let's do each side individually.

Left Side:

We have:

\frac{d}{dx}[y\sin(y)]

We can use the product rule:

(uv)'=u'v+uv'

So, our derivative is:

=\frac{d}{dx}[y]\sin(y)+y\frac{d}{dx}[\sin(y)]

We must implicitly differentiate for y. This gives us:

=\frac{dy}{dx}\sin(y)+y\frac{d}{dx}[\sin(y)]

For the sin(y), we need to use the chain rule:

u(v(x))'=u'(v(x))\cdot v'(x)

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.

So, our derivative is:

=\frac{dy}{dx}\sin(y)+y(\cos(y)\cdot\frac{dy}{dx}})

Simplify:

=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}

And we are done for the right.

Right Side:

We have:

\frac{d}{dx}[x\cos(x)]

This will be significantly easier since it's just x like normal.

Again, let's use the product rule:

=\frac{d}{dx}[x]\cos(x)+x\frac{d}{dx}[\cos(x)]

Differentiate:

=\cos(x)-x\sin(x)

So, our entire equation is:

=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}=\cos(x)-x\sin(x)

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

\frac{dy}{dx}(\sin(y)+y\cos(y))=\cos(x)-x\sin(x)

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}

And we're done!

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