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Differentiation implicit containing product and quotient. d/dm (sin y/cos x)

User Kristabel
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Answer:

To differentiate the given expression, we will need to use both the product rule and the quotient rule.

Let's first rewrite the expression using the reciprocal identity for cosine:

sin y/cos x = sin y * (cos x)^(-1)

Now we can use the product rule and the quotient rule to find the derivative with respect to m:

d/dm [sin y * (cos x)^(-1)] = sin y * d/dm[(cos x)^(-1)] + (cos x)^(-1) * d/dm[sin y]

Using the chain rule, we can find the derivative of (cos x)^(-1) with respect to m:

d/dm[(cos x)^(-1)] = -1/(cos^2 x) * (-sin x) * (dx/dm)

d/dm[(cos x)^(-1)] = sin x/cos^2 x * (dx/dm)

Using the chain rule again, we can find the derivative of sin y with respect to m:

d/dm[sin y] = cos y * (dy/dm)

Substituting these expressions back into the original equation, we get:

d/dm [sin y/cos x] = sin y * [sin x/cos^2 x * (dx/dm)] + cos y * (dy/dm) * (cos x)^(-1)

Simplifying this expression, we get:

d/dm [sin y/cos x] = (sin y * sin x/cos^2 x) * (dx/dm) + (cos y/cos x) * (dy/dm)

Therefore, the derivative of sin y/cos x with respect to m is (sin y * sin x/cos^2 x) * (dx/dm) + (cos y/cos x) * (dy/dm).

User Dtanders
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