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Calculus Rule Exercises Worked

A55-1, Find the derivative of: {$$ y= 3x^2 $$}

First solution: using rules and formulas

{$$ {dy \over dx} = {d \over dx} 3x^2 $$}

By the Constant Multiplication Rule:

{$$ {dy \over dx} = 3{d \over dx} x^2 $$}

From the Powers formula:

{$$ {d \over dx} x^2 = 2 x $$}

So,

{$$ {dy \over dx} = 3(2x) $$}

{$$ \boxed{{dy \over dx} = 6x} $$}

Second solution: by the definition:

{$$ {d \over dx} f(x) = \lim_{\Delta x\rightarrow0} {{f(x + \Delta x) - f(x)} \over {\Delta x}} $$}

{$$ {dy \over dx} = {\lim_{\Delta x\rightarrow0}} {{3(x + \Delta x)^2 - 3x^2 } \over {\Delta x}} $$}

{$$ = {\lim_{\Delta x\rightarrow0}} {{3x^2 + 6x \Delta x + 3(\Delta x)^2 - 3x^2 } \over {\Delta x}} $$}

{$$ ={ \lim_{\Delta x\rightarrow0}}{ { 6x\Delta x + 3{\Delta x}^2 } \over {\Delta x} } $$}

{$$ ={ \lim_{\Delta x\rightarrow0}}{ { (6x + 3{\Delta x}) \Delta x } \over {\Delta x} } $$}

{$$ ={ \lim_{\Delta x\rightarrow0}} { 6x + 3\Delta x } $$}

{$$ \boxed{{dy \over dx} = 6x } $$}

Third Solution: Leibniz Style

{$$ y= 3x^2 $$}

{$$ y + dy = 3(x+dx)^2 $$}

{$$ y + dy = 3x^2 + 6x \cdot dx +(dx)^2 $$}

As {$ dx $} gets smaller, {$ (dx)^2 $} is negligible:

{$$ y + dy = 3x^2 + 6x \cdot dx $$}

Since {$ y = 3x^2 $}:

{$$ 3x^2 + dy = 3x^2 + 6x \cdot dx $$}

{$$ dy = 6x \cdot dx $$}

{$$ \boxed{{dy \over dx} = 6x } $$}

Answer


{$$ \begin{gather} s = -\frac{a}{2t + 3} \cr \text{blue for positive values of a.} \end{gather} $$}

A55-17.

{$$ s = -\frac{a}{2t + 3}$$}

Solution using rules and formulas

{$$ {ds \over dt} = {d \over dt} \left(-\frac{a}{2t+3}\right) $$}

By the Constant Multiplication Rule:

{$$ {ds \over dt} = -a{d \over dt} \frac{1}{2t+3} $$}

{$$ {ds \over dt} = -a{d \over dt} (2t+3)^{-1} $$}

From the Powers formula:

{$$ {ds \over dt} = -(-1)a{d \over dt} (2t+3)^{-2} $$}

{$$ \boxed{{ds \over dt} = a(2t+3)^{-2}} $$}

Solution by the definition:

{$$ {d \over dx} f(x) = \lim_{\Delta x\rightarrow0} {{f(x + \Delta x) - f(x)} \over {\Delta x}} $$}

{$$ {d \over dt} s = \lim_{\Delta x\rightarrow0} {{{-a \over{2(t + \Delta t) +3}} - {-a \over{2t +3}}} \over {\Delta t}}$$}

{$$ {d \over dt} s = \lim_{\Delta x\rightarrow0} \left( 1 \over {\Delta t} \right) {{{-a \over{2(t + \Delta t) +3}} - {-a \over{2t +3}}}}$$}

{$$ {d \over dt} s = \lim_{\Delta x\rightarrow0} \left( 1 \over {\Delta t} \right) {{{-a(2t+3) \over(2t +3)(2(t + \Delta t) +3)} - {-a (2(t + \Delta t) +3) \over(2t +3)(2(t + \Delta t) +3)}}}$$}

{$$ {d \over dt} s = \lim_{\Delta x\rightarrow0} \left( 1 \over {\Delta t} \right) {{{-a(2t+3) +a (2(t + \Delta t) +3) \over(2t +3)(2(t + \Delta t) +3)} }}$$}

{$$ {d \over dt} s = \lim_{\Delta x\rightarrow0} \left( 1 \over {\Delta t} \right) {{{-a2t-a3 +a2t + a \Delta t +a3 \over(2t +3)(2t + 2 \Delta t +3)} }}$$}

{$$ {d \over dt} s = \lim_{\Delta x\rightarrow0} \left( 1 \over {\Delta t} \right) {{{ a \Delta t \over(2t +3)(2t + 2 \Delta t +3)} }}$$}

{$$ {d \over dt} s = \lim_{\Delta x\rightarrow0} {{{ a \over(2t +3)(2t + 2 \Delta t +3)} }}$$}

Now there is no possibility of division by zero so:

{$$ {d \over dt} s = { a \left( 1 \over(2t +3)^2 \right)} = \boxed{a(2t +3)^{-2}}$$}

Check this work. There is no textbook provided answer.


{$$ \rho = \frac{\theta^2}{1 + \theta} $$}

A55-22. Find the derivative of:

{$$ \rho = \frac{\theta^2}{1 + \theta}$$}

Solution by the definition:

{$$ {d \over dx} f(x) = \lim_{\Delta x\rightarrow0} {{f(x + \Delta x) - f(x)} \over {\Delta x}} $$}

{$$ {d \over d \theta } \rho = \lim_{\Delta \theta\rightarrow0} {{f(\theta + \Delta \theta) - f(\theta)} \over {\Delta \theta}} $$}

{$$ = \lim_{\Delta \theta\rightarrow0} {{\frac{(\theta + \Delta \theta)^2 }{1 + (\theta + \Delta \theta )} - \frac{ \theta^2}{1 + \theta}} \over {\Delta \theta}}$$}

Caution! The following algebra may be offense to sensitive viewers, even if by chance it is right.

{$$ = \lim_{\Delta \theta\rightarrow0} \left( \frac {1}{\Delta \theta} \right){{\frac{(\theta + \Delta \theta)^2 }{1 + (\theta + \Delta \theta )} - \frac{ \theta^2}{1 + \theta}}} $$}

{$$ = \lim_{\Delta \theta\rightarrow0} \left( \frac {1}{\Delta \theta} \right) {{(\theta^2 +2 \Delta \theta + (\Delta \theta)^2)(1 + \theta) - \theta^2(1 + \theta + \Delta \theta )} \over (1 + \theta + \Delta \theta )(1 + \theta)} = $$}

{$$ \lim_{\Delta \theta\rightarrow0} \left( \frac {1}{\Delta \theta} \right) {{\theta^2 +2 \theta(\Delta \theta) + (\Delta \theta)^2 + \theta^3 + 2 \theta^2 (\Delta \theta) + \theta(\Delta \theta)^2 - \theta^2 - \theta^3 - (\Delta \theta)\theta^2 )} \over (1 + \theta + \Delta \theta )(1 + \theta)} $$}

{$$ = \lim_{\Delta \theta\rightarrow0} \left( \frac {1}{\Delta \theta} \right) {{2 \theta(\Delta \theta) + (\Delta \theta)^2 + 2 \theta^2 (\Delta \theta) + \theta(\Delta \theta)^2 - (\Delta \theta)\theta^2 )} \over (1 + \theta + \Delta \theta )(1 + \theta)} $$}

{$$ = \lim_{\Delta \theta\rightarrow0} {{2 \theta + (\Delta \theta) + 2 \theta^2 + \theta(\Delta \theta) - \theta^2 )} \over (1 + \theta + \Delta \theta )(1 + \theta)} $$}

The threat of division by zero being removed, the limit is found by substituting {$ 0 \text{ for } \Delta \theta $}.

{$$ \boxed{{d \over d \theta } \rho = {{2 \theta + \theta^2 )} \over {(1 + \theta)^2}}} $$}

Solution by the Product Rule

{$$ {d \over dx}(f(x)g(x)) = g(x){d \over dx}f(x) + f(x){d \over dx}g(x) \tag{Product Rule}$$}

{$$ \text{Let } f(\theta) = \theta^2 \text{ and } {g(\theta) = } {{1} \over (1 + \theta) } = (1 + \theta)^{-1} \text{.} $$}

{$$ {d \over d \theta }(f(\theta)g(\theta)) = g(\theta){d \over d\theta}f(\theta) + f(\theta){d \over d\theta}g(\theta) $$}

{$$ {d \over d \theta } \left( \frac{\theta^2}{1 + \theta} \right) = (1 + \theta)^{-1}{d \over d\theta} \theta^2 + \theta^2 {d \over d\theta} (1 + \theta)^{-1} $$}

Applying the Powers formula:

{$$ {d \over d \theta } \rho = (1 + \theta)^{-1} 2 \theta + \theta^2 (-1) (1 + \theta)^{-2} $$}

Which is the answer, but let's put some lipstick on this pig:

{$$ {d \over d \theta } \rho = {{2 \theta} \over (1 + \theta)} - {{ \theta^2} \over { (1 + \theta)^{2}}} $$}

{$$ {d \over d \theta } \rho = { {{2 \theta} + 2 \theta^2 - \theta^2} \over { (1 + \theta)^{2}}} $$}

{$$ \boxed {{d \over d \theta } \rho = { {{2 \theta} + \theta^2} \over { (1 + \theta)^{2}}}} $$}

Solution by the Quotient Rule

{$$ {d \over dx} \left({{f(x)} \over {g(x)}}\right) = { {g(x){d \over dx}f(x) - f(x){d \over dx}g(x)} \over {[g(x)]^2}} \tag{Quotient Rule} $$}

{$$ \text{Let } f(\theta) = \theta^2 \text{ and } g(\theta) = (1 + \theta) \text{.} $$}

{$$ {d \over d \theta} \left({{f(\theta)} \over {g(\theta)}}\right) = { {g(\theta){d \over d \theta}f(\theta) - f(\theta){d \over d \theta}g(\theta)} \over {[g(\theta)]^2}} $$}

{$$ {d \over d \theta} \rho = { {(1+\theta){d \over d \theta}(\theta^2) - \theta^2 {d \over d \theta}(1+ \theta) } \over {(1+\theta)^2}} $$}

{$$ {d \over d \theta} \rho = { {(1+\theta)2(\theta) - \theta^2 (1) } \over {(1+\theta)^2}} $$}

{$$ {d \over d \theta} \rho = { {2\theta + 2\theta^2 - \theta^2 } \over {(1+\theta)^2}} $$}

{$$ \boxed {{d \over d \theta } \rho = { {{2 \theta} + \theta^2} \over { (1 + \theta)^{2}}}} $$}

Check this work. There is no textbook provided answer.


{$$ y = x^{13} $$}

B24-1. Find the derivative of:

{$$ y = x^{13} $$}

Of course real-world people do this by inspection in light of the powers formula. But in Math world:

Solution by the definition:

{$$ {d \over dx} f(x) = \lim_{\Delta x\rightarrow0} {{f(x + \Delta x) - f(x)} \over {\Delta x}} $$}

{$$ {d \over dx} y = \lim_{\Delta x\rightarrow0} {{(x + \Delta x)^{13} - x^{13}} \over {\Delta x}} $$}

Moving the denominator out of the way and invoking the Binomial Theorem (the action is in the summation variable k) we peel the first term off the sum and use it to cancel the negative term:

{$$ ={ \lim_{\Delta x\rightarrow0} \left( \frac {1}{\Delta \theta} \right) \left( \sum_{k=0}^{13} {13 \choose k}x^{13-k} (\Delta x)^k - x^{13} \right)} $$}

{$$ ={ \lim_{\Delta x\rightarrow0} \left( \frac {1}{\Delta \theta} \right) \left( x^{13} + \sum_{k=1}^{13} {13 \choose k}x^{13-k} (\Delta x)^k - x^{13} \right)} $$}

{$$ ={ \lim_{\Delta x\rightarrow0} \left( \frac {1}{\Delta \theta} \right) \left( \sum_{k=1}^{13} {13 \choose k}x^{13-k} (\Delta x)^k \right)} $$}

Now every term of the sum contains a factor of {$ \Delta x $} so we can perform the division:

{$$ ={ \lim_{\Delta x\rightarrow0} \left( \sum_{k=1}^{13} {13 \choose k}x^{13-k} (\Delta x)^{k-1} \right)} $$}

{$$ ={ \lim_{\Delta x\rightarrow0}\left( {13 \choose 1}x^{13-1}(\Delta x)^0 + \sum_{k=2}^{13} {13 \choose k}x^{13-k} (\Delta x)^{k-1} \right)} $$}

{$$ {d \over dx} y={ \lim_{\Delta x\rightarrow0}\left( \frac {13!} {1!12!}x^{12} + \sum_{k=2}^{13} {13 \choose k}x^{13-k} (\Delta x)^{k-1} \right)} $$}

But the sum now contains non-zero powers of {$ \Delta x $} which go to zero. So:

{$$ \boxed {{d \over dx} y= 13x^{12} } $$}


{$$ z = \sqrt[3] u $$}

B25-5. Find the derivative of:

{$$ z = \sqrt[3] u $$}

Solution by the powers formula

{$$ z = \sqrt[3] u = u^{\frac 1 3} $$}

by inspection:

{$$ \boxed {\dfrac{d}{du} z = { 1 \over 3 } u^{\bf- {\frac 2 3}}} $$}

Solution by the definition:

Definition:

{$$ {d \over dx} f(x) = \lim_{\Delta x\rightarrow0} {{f(x + \Delta x) -f(x)} \over {\Delta x}} $$}

Plugging in:

{$$ {d \over dx} z = \lim_{\Delta u\rightarrow0} {{(u + \Delta u)^{\frac 1 3} - u^{\frac 1 3} \over {\Delta u}} } $$}

Now for some hairy algebra:

{$$ = \lim_{\Delta u\rightarrow0} {{[(u + \Delta u)^{\frac 1 3} - u^{\frac 1 3}][(u + \Delta u)^{\frac 2 3} + (u + \Delta u)^{\frac 1 3}u^{\frac 1 3} + u^{\frac 2 3}]} \over {{\Delta u}[(u + \Delta u)^{\frac 2 3} + (u + \Delta u)^{\frac 1 3}u^{\frac 1 3} + u^{\frac 2 3}]}} $$}

{$$ = \lim_{\Delta u\rightarrow0} {{(u+ \Delta u) + (u + \Delta u)^{\frac 2 3}u^{\frac 1 3} - (u + \Delta u)^{\frac 2 3}u^{\frac 1 3} + (u + \Delta u)^{\frac 1 3}u^{\frac 2 3} - (u + \Delta u)^{\frac 1 3}u^{\frac 2 3} - u} \over {{\Delta u}[(u + \Delta u)^{\frac 2 3} + (u + \Delta u)^{\frac 1 3}u^{\frac 1 3} + u^{\frac 2 3}]}} $$}

{$$ = \lim_{\Delta u\rightarrow0} {{ \Delta u} \over {{\Delta u}[(u + \Delta u)^{\frac 2 3} + (u + \Delta u)^{\frac 1 3}u^{\frac 1 3} + u^{\frac 2 3}]}} $$}

Factor the {$ \Delta u $} from the numerator and denominator:

{$$ = \lim_{\Delta u\rightarrow0} {{ 1 }\over {(u + \Delta u)^{\frac 2 3} + (u + \Delta u)^{\frac 1 3}u^{\frac 1 3} + u^{\frac 2 3}}} $$}

Now we can take the limit:

{$$ \dfrac {d} {du} z = {{ 1} \over {u^{\frac 2 3} + u^{\frac 1 3}u^{\frac 1 3} + u^{\frac 2 3}}} $$}

{$$ \dfrac {d} {du} z = {{ 1} \over {u^{\frac 2 3} + u^{\frac 2 3} + u^{\frac 2 3}}} $$}

{$$ \dfrac {d} {du} z = {{ 1 }\over { 3u^{\frac 2 3}}} $$}

{$$ \boxed {\dfrac{d}{du} z = { 1 \over 3 } u^{\bf- {\frac 2 3}}} $$}


Sources:

  1. Elements of the Differential and Integral Calculus Wikisource
  2. Calculus Made Easy by Silvanus P. Thompson Project Gutenberg:

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August 05, 2017

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