Exam focus: 1. @ f(x) = -sqrtx+1 2. A body moves in a straight line according to the law of motion S = t3 − 2t2 + t. Find velocity and acceleration after 5 seconds. lim x->5 (x2+1)/(x2-x) lim x->2 (x2-4)/(x-2) * If Z = x2 + y2 + x2y3, 1. dZ/dx 2. dZ/dy if x=1, y=1

Here are the solutions to the problems from the image:

(1). Differentiate $f(x) = -\sqrt{x+1}$

Step 1: Rewrite the function using exponential notation.

$$f(x) = -(x+1)^{1/2}$$

Step 2: Apply the chain rule. Let $u = x+1$, so $\frac{du}{dx} = 1$.

$$\frac{df}{dx} = -\frac{1}{2}(x+1)^{(1/2)-1} \cdot \frac{d}{dx}(x+1)$$

Step 3: Simplify the expression.

$$\frac{df}{dx} = -\frac{1}{2}(x+1)^{-1/2} \cdot 1$$ $$\frac{df}{dx} = \boxed{-\frac{1}{2\sqrt{x+1}}}$$

(2). A body moves in a straight line according to the law of motion $S = t^3 - 2t^2 + t$. Find velocity and acceleration after 5 seconds.

Step 1: Find the velocity function $V(t)$ by differentiating the displacement function $S(t)$ with respect to $t$.

$$V(t) = \frac{dS}{dt} = \frac{d}{dt}(t^3 - 2t^2 + t) = 3t^2 - 4t + 1$$

Step 2: Calculate the velocity at $t=5$ seconds.

$$V(5) = 3(5)^2 - 4(5) + 1 = 3(25) - 20 + 1 = 75 - 20 + 1 = \boxed{56 \text{ units/s}}$$

Step 3: Find the acceleration function $A(t)$ by differentiating the velocity function $V(t)$ with respect to $t$.

$$A(t) = \frac{dV}{dt} = \frac{d}{dt}(3t^2 - 4t + 1) = 6t - 4$$

Step 4: Calculate the acceleration at $t=5$ seconds.

$$A(5) = 6(5) - 4 = 30 - 4 = \boxed{26 \text{ units/s}^2}$$

(Lim). Find $\lim_{x \to 5} \frac{x^2 + 1}{x^2 - x}$

Step 1: Substitute $x=5$ into the expression.

$$\lim_{x \to 5} \frac{x^2 + 1}{x^2 - x} = \frac{(5)^2 + 1}{(5)^2 - 5}$$

Step 2: Evaluate the numerator and the denominator.

$$= \frac{25 + 1}{25 - 5} = \frac{26}{20}$$

Step 3: Simplify the fraction.

$$= \boxed{\frac{13}{10}}$$

(Lim). Find $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Step 1: Substitute $x=2$ into the expression. This results in $\frac{0}{0}$, an indeterminate form.

$$\frac{(2)^2 - 4}{2 - 2} = \frac{4 - 4}{0} = \frac{0}{0}$$

Step 2: Factor the numerator using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$).

$$x^2 - 4 = (x-2)(x+2)$$

Step 3: Rewrite the limit expression and cancel out the common factor $(x-2)$.

$$\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2)$$

Step 4: Substitute $x=2$ into the simplified expression.

$$= 2 + 2 = \boxed{4}$$

(If Z). If $Z = x^2 + y^2 + xy^3$, find (i) $\frac{\partial Z}{\partial x}$ and (ii) $\frac{\partial Z}{\partial y}$ if $x=1, y=1$.