Hello Noora,
a) Let's define:
$$f(x)=x^{\frac{4}{3}}$$
Hence:
$$f'(x)=\frac{4}{3}x^{\frac{1}{3}}$$
For a small $\Delta x$, we know:
$$\frac{\Delta f}{\Delta x}\approx\frac{df}{dx}$$
Multiplying through by $\Delta x$ and using $Delta f=f\left(x+\Delta x \right)-f(x)$ we have:
$$f\left(x+\Delta x \right)-f(x)=\frac{df}{dx}\Delta x$$
$$f\left(x+\Delta x \right)=\frac{df}{dx}\Delta x+f(x)$$
Now, choosing:
$$x=27,\,\Delta x=-0.02$$
and using our function definition, we obtain:
$$(26.98)^{\frac{4}{3}}\approx\frac{4}{3}(27)^{\frac{1}{3}}(-0.02)+(27)^{\frac{4}{3}}=81-0.08=80.92$$
b) We are given the implicit relation:
$$x^2y^2-3y=2x^4-4$$
Implicitly differentiating with respect to $x$, we find:
$$x^2\cdot2y\frac{dy}{dx}+2xy^2-3\frac{dy}{dx}=8x^3$$
$$\frac{dy}{dx}\left(2x^2y-3 \right)=8x^3-2xy^2$$
$$\frac{dy}{dx}=\frac{2x\left(4x^2-y^2 \right)}{2x^2y-3}$$
Now, for a small $\Delta x$, we have:
$$\frac{\Delta y}{\Delta x}\approx\frac{dy}{dx}$$
$$y\left(x+\Delta x \right)\approx\frac{dy}{dx}\cdot\Delta x+y(x)$$
Using the given:
$$(x,y)=(1,1),\,\Delta x=0.1$$
we find:
$$y(1.1)\approx\frac{2(1)\left(4(1)^2-(1)^2 \right)}{2(1)^2(1)-3}\cdot0.1+1=0.4$$
