Tutor profile: Daniel C.

Find $$f'(x)$$ of the function: $(f(x) = 2 \cos^2(\sin(x))$).

To solve this problem, we need to use the chain rule: $((f(u(x)))'= f'(u(x)) \cdot u'(x)$). Here, $$u= \sin(x)$$, $$f= 2 \cos^2(u)$$. This is complicated by the fact that the derivative of $$f$$ with respect to $$u$$ requires us to use the chain rule yet again. Let's start then by focusing on this term. Let's substitute again. We want the derivative of $$f$$ with respect to $$u$$. To get that, we're using the chain rule, so we should define $$w=\cos(u)$$ so that $$f = 2 w^2$$. Then, the chain rule tells us $(f(w(u))'= f'(w(u)) \cdot w'(u)$). Note that I'm just rewriting the chain rule with different variables. Then, the first term is just the derivative of $$2w^2$$ and the second term is just the derivative of $$\cos(u)$$: $(f(w(u))'= 2 \cdot (2 w) \cdot (-\sin (u))$). Since $$w= \cos(u)$$, this simplifies to give: $(f'(u) = -4\sin(u)\cos(u)$). Don't forget that $$u=\sin(x)$$, so we can substitute that to get: $(f'(u(x)) = -4\sin(\sin(x))\cos(\sin(x))$). That was the hard part! We did all that work to get the first term in $((f(u(x)))'= f'(u(x)) \cdot u'(x)$). Now we just have to take the derivative of the second term in the chain rule, which is simply the derivative of $$\sin(x)$$. So, we have: $(u'(x) = \cos(x)$). Putting this all together, the final answer is: $(f'(x) = -4\sin(\sin(x))\cos(\sin(x)) \cos(x)$).

A variable force $$\mathbf{𝐹}$$ =2 𝑥 $$\hat{i}$$+3 $$\hat{j}$$ is applied to an object when it is located at the origin. While this force is applied, it is displaced by 3 m along the $$\textit{x}$$ direction. How much work is done? What is its final velocity if it was initially moving with a velocity of 2 $$\textit{m/s}$$ along the $$\textit{x}$$ direction? Suppose its mass is 2 kg.

This problem requires a variable force, so integration is required. The work done is defined as: $(W = \int_{(x_{i}, y_{i}, z_{i})}^{(x_{f}, y_{f}, z_{f})} \mathbf{F} \cdot \mathbf{dr}$). Since the displacement is only along the $$\textit{x}$$-direction, only that component of the force does work, so we can rewrite this as: $(W = \int_{x_{i}}^{x_{f}} F_{x} dx$). Substituting into this expression, where the initial position $$x_{i}=0$$ and the final position $$x_{f}=3 $$, we get: $(W = \int_{0}^{3} 2x dx$). Integrating: $(W = x^2|_{0}^{3}= 3^2-0^2=9 J$). We know that 9 $$J$$ of work are done by this variable force. We can now use the work-kinetic energy theorem to obtain the final velocity: $(W =\Delta KE$). where we are taking the difference between the final and initial kinetic energies: $(W =\frac{1}{2} mv_{f}^2-\frac{1}{2} mv_{i}^2 $). Since we know the work and we know the object's initial velocity, we just have to plug in and solve for $$v_{f}$$: $(9J =\frac{1}{2} (2 kg)v_{f}^2-\frac{1}{2} (2 kg)(2 m/s)^2 $). $(9J =v_{f}^2-4 $). $(v_{f} =\sqrt{13}$). Therefore, the final velocity is $$3.61 m/s$$.

Solve the following equation: $(|2 x -3| - 5 < 4$).

This is an absolute value expression, so we need to be careful with how we deal with it. Before we do that though, we should isolate the absolute value on one side of the equation by adding 5 to both sides: $(|2 x -3| -5+5 < 4+5$). $(|2 x -3|< 9$). This expression has two solutions. The first solution can be found by dropping the absolute value: $(2 x -3< 9$). Solve for $$x$$ by adding3 to both sides: $(2 x -3+3< 9+3$). $(2 x < 12$). Then, we can isolate $$x$$ by dividing by 2 on both sides: $(\frac{2 x}{2} < \frac{12}{2}$). $( x < 6$) The second solution can be found by dropping the absolute value and then negating the expression on the left hand side: $(-(2 x -3)< 9$). Using the distributive property, this becomes: $(-2 x +3< 9$). Solve for $$x$$ by subtracting 3 from both sides: $(-2 x +3-3< 9-3$). $(-2 x < 6$). Then, isolate $$x$$ by dividing by -2. Don't forget that when you divide by a negative number, you have to flip the less than sign to a greater than sign: $(\frac{-2 x}{-2} < \frac{6}{-2}$). $(x> -3$). Putting the two answers together: $(-3<x <6$).