91Ó°ÊÓ

To find the partial derivative of \(f(x, y) = x^{2y}\) with respect to \(x\), we will use the chain rule. Recall that the chain rule for a function of the form \(f(x) = u(v(x))\) is given by \(f'(x) = u'(v(x)) * v'(x)\). In our case, let \(u(x) = x^{2y}\) and \(v(x) = 2y\). We have: $$\frac{\partial f}{\partial x} = \frac{\partial (x^{2y})}{\partial x}$$ Using the chain rule, we get: $$\frac{\partial f}{\partial x} = (2y) x^{(2y-1)}$$ So, the partial derivative of \(f(x, y)\) with respect to \(x\) is: $$\frac{\partial f}{\partial x} = 2yx^{(2y-1)}$$

To find the partial derivative of \(f(x, y) = x^{2y}\) with respect to \(y\), we will first rewrite the function as an exponential function with base \(e\) using the identity \(a^b = e^{(\ln a)b}\): $$f(x, y) = x^{2y} = e^{(\ln x)(2y)}$$ Now, find the partial derivative with respect to \(y\), using the chain rule: $$\frac{\partial f}{\partial y} = \frac{\partial (e^{(\ln x)(2y)})}{\partial y}$$ The derivative of \(e^u\) with respect to \(u\) is \(e^u\). Using the chain rule, we have: $$\frac{\partial f}{\partial y} = e^{(\ln x)(2y)}\cdot \frac{\partial((\ln x)(2y))}{\partial y}$$ Now differentiate \((\ln x)(2y)\) with respect to \(y\): $$\frac{\partial((\ln x)(2y))}{\partial y} = 2\ln x$$ So, the partial derivative of \(f(x, y)\) with respect to \(y\) is: $$\frac{\partial f}{\partial y} = 2\ln x\cdot e^{(\ln x)(2y)} = 2\ln x\cdot x^{2y}$$ Now we have both first partial derivatives: $$\frac{\partial f}{\partial x} = 2yx^{(2y-1)}$$ $$\frac{\partial f}{\partial y} = 2\ln x\cdot x^{2y}$$