More Mixed-Rule Derivative Practice
Instructions. For each function, find \(f'(x)\) without simplifying.
Assume \(y=y(x)\) (implicit differentiation when needed).
Click to reveal each step of the structured solution.
Challenge Q1
\[ f_1(x)=\frac{(x+y^2)\,e^{\cos(xy)}}{1+x^2}. \]
Step 1
\[ N(x)=(x+y^2)e^{\cos(xy)},\qquad D(x)=1+x^2. \] \[ f_1'(x)=\frac{D(x)N'(x)-N(x)D'(x)}{[D(x)]^2}. \]Step 2
\[ A(x)=x+y^2,\quad A'(x)=1+2y\,y',\qquad B(x)=e^{\cos(xy)},\quad B'(x)=e^{\cos(xy)}(-\sin(xy))(y+x\,y'). \] \[ N'(x)=(1+2y\,y')e^{\cos(xy)}+(x+y^2)e^{\cos(xy)}[-\sin(xy)](y+x\,y'). \]Step 3
\[ D'(x)=2x. \]Step 4
\[ \boxed{f_1'(x)= \frac{(1+x^2)\!\left[(1+2y\,y')e^{\cos(xy)}+(x+y^2)e^{\cos(xy)}[-\sin(xy)](y+x\,y')\right] -2x(x+y^2)e^{\cos(xy)}}{(1+x^2)^2}}. \]Challenge Q2
\[ f_2(x)=(x^2+y)\,\sin(\ln(1+xy)). \]
Step 1
\[ f_2(x)=A(x)\,B(x),\quad A(x)=x^2+y,\; B(x)=\sin(\ln(1+xy)). \] \[ f_2'(x)=A'(x)B(x)+A(x)B'(x). \]Step 2
\[ A'(x)=2x+y'. \] \[ B'(x)=\cos(\ln(1+xy))\cdot\frac{1}{1+xy}\cdot(y+x\,y') =\frac{\cos(\ln(1+xy))(y+x\,y')}{1+xy}. \]Step 3
\[ \boxed{f_2'(x)=(2x+y')\sin(\ln(1+xy)) +(x^2+y)\frac{\cos(\ln(1+xy))(y+x\,y')}{1+xy}}. \]Challenge Q3
\[ f_3(x)=\frac{\ln(x+y)\,e^{xy}}{\cos y}. \]
Step 1
\[ N(x)=\ln(x+y)e^{xy},\quad D(x)=\cos y. \] \[ f_3'(x)=\frac{D(x)N'(x)-N(x)D'(x)}{[D(x)]^2}. \]Step 2
\[ A(x)=\ln(x+y),\; A'(x)=\frac{1+y'}{x+y},\quad B(x)=e^{xy},\; B'(x)=e^{xy}(y+x\,y'). \] \[ N'(x)=A'(x)B(x)+A(x)B'(x) =\frac{1+y'}{x+y}e^{xy}+\ln(x+y)e^{xy}(y+x\,y'). \]Step 3
\[ D'(x)=-\sin y\,y'. \]Step 4
\[ \boxed{f_3'(x)= \frac{\cos y\!\left[\frac{1+y'}{x+y}e^{xy}+\ln(x+y)e^{xy}(y+x\,y')\right] +\sin y\,y'\ln(x+y)e^{xy}}{(\cos y)^2}}. \]Challenge Q4
\[ f_4(x)=e^{y}\cdot\frac{x+\sin(xy)}{1+y^2}. \]
Step 1
\[ f_4(x)=P(x)\,Q(x),\quad P(x)=e^{y},\quad Q(x)=\frac{x+\sin(xy)}{1+y^2}. \] \[ f_4'(x)=P'(x)Q(x)+P(x)Q'(x). \]Step 2
\[ P'(x)=e^{y}y'. \] \[ Q(x)=\frac{R(x)}{S(x)},\quad R(x)=x+\sin(xy),\; S(x)=1+y^2. \] \[ R'(x)=1+\cos(xy)(y+x\,y'),\quad S'(x)=2y\,y'. \] \[ Q'(x)=\frac{R'S-RS'}{S^2} =\frac{(1+\cos(xy)(y+x\,y'))(1+y^2)-(x+\sin(xy))(2y\,y')}{(1+y^2)^2}. \]Step 3
\[ \boxed{f_4'(x)= e^{y}y'\frac{x+\sin(xy)}{1+y^2} +e^{y}\frac{(1+\cos(xy)(y+x\,y'))(1+y^2)-(x+\sin(xy))(2y\,y')}{(1+y^2)^2}}. \]Challenge Q5
\[ f_5(x)=\frac{(x^2+1)\tan(xy)}{e^{x+y}}. \]