Logarithmic implicit differentiation $x^{\sin(y)} = y^{\cos(x)}$

Question

The question I'm stuck on is:

Let $y = f(x)$ be implicitly defined by $x^{\sin(y)} = y^{\cos(x)}$ .Compute y' in terms of x and y. (HINT Can logarithms help you?)

I was thinking of just using the normal power and product rules, but I thought there may be a quicker way that I wasn't thinking of.

Thanks

Answer 1

First take the logs of both sides to obtain $\sin(y)\ln(x)=\cos(x)\ln(y)$.

Now you can implicitly differentiate using the product and chain rule to obtain:

$\frac{\sin(y)}{x}+\ln(x)\cos(y)\frac{dy}{dx}=-\sin(x)\ln(y)+\cos(x)\frac{1}{y}\frac{dy}{dx}$ and now re-arranging we obtain:

$$\frac{dy}{dx}=\frac{-\frac{\sin(y)}{x}-\sin(x)\ln(y)}{\ln(x)\cos(y)-\frac{\cos(x)}{y}}$$ and multiplying by $xy$ gives

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

Answer 2

Question: find $\large{\frac{dy}{dx}}$ for $\large{x^{\sin y}=y^{\cos x}}$

Let, $$\large{f\left(x,y\right)=x^{\sin y}-y^{\cos x}}$$ then ,$$\frac{\partial}{\partial x}f\left(x,y\right)=\sin\left(y\right)\left(\frac{x^{\sin y}}{x}\right)+y^{\cos x}\ln\left(y\right)\left(\sin\left(x\right)\right)$$

$$\text{and}$$

$$\frac{\partial}{\partial y}f\left(x,y\right)=x^{\sin y}\cos\left(y\right)\ln\left(x\right)-\frac{\left(\cos x\right)y^{\cos x}}{y}$$

since, $$\large{-\frac{\frac{\partial}{\partial x}f\left(x,y\right)}{\frac{\partial}{\partial y}f\left(x,y\right)}=\frac{d}{dx}y}$$

Hence the answer,

$$\large{\frac{dy}{dx}=-\frac{\sin\left(y\right)\left(\frac{x^{\sin y}}{x}\right)+y^{\cos x}\ln\left(y\right)\left(\sin\left(x\right)\right)}{x^{\sin y}\cos\left(y\right)\ln\left(x\right)-\frac{\left(\cos x\right)y^{\cos x}}{y}}}$$

Or, $$\large \frac{dy}{dx}={\frac{\sin\left(y\right)\left(x^{\sin y}\right)+x\cdot y^{\cos x}\ln\left(y\right)\left(\sin\left(x\right)\right)}{\left(\cos x\right)y^{\cos x}-y\cdot x^{\sin y}\cos\left(y\right)\ln\left(x\right)}\cdot\frac{y}{x}}$$

Using the first equation given to us, we can simplify the above in $$\large{\frac{dy}{dx}=\frac{\sin\left(y\right)+x\cdot\ln\left(y\right)\left(\sin\left(x\right)\right)}{\left(\cos x\right)-y\cdot\cos\left(y\right)\ln\left(x\right)}\cdot\frac{y}{x}}$$

Answer 3

Just as a note, there is actually a general formula for $q^r$ (I used $q$ and $r$ instead of $u$ and $v$ because the current math font makes $u$ and $v$ look too similar), though almost no one teaches it anymore. If you look at it you will see that it is a combination of the power and exponent rules:

$$d(q^r) = rq^{r-1}dq + ln(q)q^rdr$$