MA005: Calculus I

You definitely need to memorize the differentiation rules, but it is vitally important that you also know how to use them. Sometimes it is clear that the function we want to differentiate is a sum or product of two obvious functions, but we commonly need to differentiate functions which involve several operations and functions. Memorizing the differentiation rules is only the first step in learning to use them. 

Example 7: Calculate $\mathbf{D}\left(\mathrm{x}^{5}+\mathrm{x} \cdot \sin (\mathrm{x})\right)$

Solution: This function is more difficult because it involves both an addition and a multiplication. Which rule(s) should we use, or, more importantly, which rule should we use first?

$\mathbf{D}\left(\mathrm{x}^{5}+\mathrm{x} \cdot \sin (\mathrm{x})\right) \quad=\mathrm{D}\left(\mathrm{x}^{5}\right)+\mathbf{D}(\mathrm{x} \cdot \sin (\mathrm{x}))$

applying the Sum Rule and trading one derivative for two easier ones

$=5 x^{4}+\{x \cdot D(\sin (x))+\sin (x) \cdot D(x)\}$ applying the product rule to $D( x.sin(x) )$ 

$=5x4 + x.cos(x) + sin(x)$ this expression has no more derivatives so we are done.

If you were evaluating the function $x^5 + x sin(x)$ for some particular value of $x$, you would (1) raise $x$ to the 5th power, (2) calculate $sin(x)$, (3) multiply $sin(x)$ by $x$, and (4) your FINAL evaluation step, SUM the values of  $x^5$ and $x sin(x)$.  

The FINAL step of your evaluation of $f$ indicates the FIRST rule to use to calculate the derivative of $f$. 

Practice 8: Which differentiation rule should you apply FIRST for each of the following: 

(a) $x \cdot \cos (x)-x^{3} \cdot \sin (x)$

(b) $(2 x-3) \cdot \cos (x)$

(c) $2 \cos (\mathrm{x})-7 \mathrm{x}^{2}$

(d) $\frac{\cos (x)+3 x}{\sqrt{x}}$

Practice 9: Calculate $D( \left.\frac{\mathrm{x}^{2}-5}{\sin (\mathrm{x})}\right)$ and $\frac{\mathbf{d}}{\mathbf{d t}}\left(\frac{\mathrm{t}^{2}-5}{\mathrm{t} \cdot \sin (\mathrm{t})}\right)$.

Example 8: A weight attached to a spring is oscillating up and down.

Over a period of time, the motion becomes "damped" because of friction and air resistance (Fig. 5), and its height at time $\mathrm{t}$ seconds is $\mathrm{h}(\mathrm{t})=5+\frac{\sin (\mathrm{t})}{1+\mathrm{t}}$ feet

What are the height and velocity of the weight after 2 seconds?

Solution: The height is

$\mathrm{h}(2)=5+\frac{\sin (2)}{1+2}=5+\frac{.909}{3}=5.303$ feet above the ground.

The velocity is $\mathrm{h}^{\prime}(2)$.

$\mathrm{h}^{\prime}(\mathrm{t})=0+\frac{(1+\mathrm{t}) \cdot \mathbf{D}(\sin (\mathrm{t}))-\sin (\mathrm{t}) \cdot \mathbf{D}(1+\mathrm{t})}{(1+\mathrm{t})^{2}}=\frac{(1+\mathrm{t}) \cdot \cos (\mathrm{t})-\sin (\mathrm{t})}{(1+\mathrm{t})^{2}}$

so $h^{\prime}(2)=\frac{3 \cos (2)-\sin (2)}{9}=\frac{-2.158}{9} \approx-0.24$ feet per second.

Practice 10: What are the height and velocity of the weight in the previous example after 5 seconds? What are the height and velocity of the weight be after a "long time"? 

Example 9: Calculate $D( x.sin(x).cos(x) )$. 

Solution: Clearly we need to use the Product Rule since the only operation in this function is multiplication, but the Product Rule deals with a product of two functions and we have the product of three; $x$ and $sin(x)$ and $cos(x)$. However, if we think of our two functions as $f(x) = x.sin(x)$ and $g(x) = cos(x)$, then we do have the product of two functions and 

$D( x.sin(x).cos(x) ) = D( f(x).g(x) ) = f(x).D( g(x) ) + g(x).D( f(x) )$

We are not done, but we have traded one hard derivative for two easier ones. We know that $D( cos(x) ) = –sin(x)$, and we can use the Product Rule (again) to calculate $D( x sin(x) )$. Then the last line of our calculation becomes 

$= –x sin2(x) + cos(x) { x cos(x) + sin(x) (1)} = –x sin2(x) + x cos2(x) + cos(x) sin(x)$.