The derivative of the function is equal to

\(\displaystyle f'=-2x\gamma \sin(x^2)\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

The constant may seem intimidating, but we treat it as another constant!

To find the derivative of \(\displaystyle \delta\) with respect to x, we must differentiate both sides of the equation with respect to x:

\(\displaystyle 0=2\delta xe^{x^2}+e^{x^2}\frac{\mathrm{d} \delta}{\mathrm{d} x}\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Solving for \(\displaystyle \frac{\mathrm{d} \delta}{\mathrm{d} x}\), we get

\(\displaystyle \frac{\mathrm{d} \delta}{\mathrm{d} x}=\frac{-2\delta x e^{x^2}}{e^{x^2}}=-2\delta x\)

Note that the chain rule was used because of the exponential and because \(\displaystyle \delta\) is a function of x.

The derivative of the function is equal to

\(\displaystyle f'=2x\sec(x^2)\tan(x^2)+\frac{4x}{2x^2}=2x\sec(x^2)\tan(x^2)+\frac{2}{x}\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \ln(x)=\frac{1}{x}\)

Note that the chain rule was used on the secant function as well as the natural logarithm function. 

To determine \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}\), we must take the derivative of both sides of the equation with respect to x:

\(\displaystyle e^y+2xe^y \frac{\mathrm{d} y}{\mathrm{d} x}= 2y\frac{\mathrm{d} y}{\mathrm{d} x}\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x\)

Rearranging and solving for \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}\), we get

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{e^y}{2y-2xe^y}\)

Differentiate both sides with respect to \(\displaystyle x\) using the chain rule and the product rule as:

\(\displaystyle \left ((1)(y^2) + (x)(2y)\left (\frac{dy}{dx} \right ) \right )+\left ((2x)(y)+(x^2)\left(\frac{dy}{dx} \right ) \right )-1=0\)

Then solve for \(\displaystyle dy/dx\) as if it were our unknown:

\(\displaystyle \left (\frac{dy}{dx} \right )(2xy+x^2)=1-2xy-y^2\)

\(\displaystyle \frac{dy}{dx}=\frac{1-2xy-y^2}{2xy+x^2}\).

Finally, evaluate \(\displaystyle dy/dx\) at the point \(\displaystyle (1,-3)\) to obtain the slope through that point:

\(\displaystyle \frac{1-2(1)(-3)-(-3)^2}{2(1)(-3)+(1)^2} = \frac{-2}{-5}=\frac{2}{5}\).

Differentiate both sides of the equation with respect to \(\displaystyle x\), using the chain rule on the \(\displaystyle y^4\) term:

\(\displaystyle x^4+y^4=R^4\)

\(\displaystyle 4x^3+4y^3\left (\frac{dy}{dx} \right )=0\)

Then solve for \(\displaystyle \frac{dy}{dx}\) as if it were our unknown:

\(\displaystyle \frac{dy}{dx}=\frac{-4x^3}{4y^3}=-\frac{x^3}{y^3}\).

Comparing this to the figure, our answer makes sense, because the slope of the squircle is \(\displaystyle 0\) wherever \(\displaystyle x=0\) (as it crosses the \(\displaystyle y\) axis) and undefined (vertical) wherever \(\displaystyle y=0\) (as it crosses the \(\displaystyle x\) axis). Lastly, we note that in the first quadrant (where \(\displaystyle x>0\) and \(\displaystyle y>0\)), the slope of the squircle is negative, which is exactly what we observe in the figure.