Maria's copy of Differentiation using the product rule

Number of Questions:	11
Marks Available:	44
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Question 1

$\simplify[std]{f(x) = x ^ {m} * ({a} * x^2+{b})^{n}}$

Differentiate $f(x)$.

The answer can be written in the form

\[\frac{\mathrm{d}f}{\mathrm{d}x}=\simplify[std]{x^{m-1}({a}x^2+{b})^{n-1}*g(x)}\] for a polynomial $g(x)$.

You have to find $g(x)$.

$g(x)=$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{136 x^{ 2 } - 9}$

Click on Show steps for more information.

You should use the the product rule and the chain rule for this example.

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then

\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[std]{u = x ^ {m}}\Rightarrow \simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\]

\[\simplify[std]{v = ({a} * x^2+{b})^{n}} \Rightarrow \simplify[std]{Diff(v,x,1) = {2*n*a}*x * ({a} * x^2+{b})^{n-1}}\]

For this last differentiation we used the chain rule.

Hence on substituting into the product rule above we get:

\begin{align}
\frac{\mathrm{d}f}{\mathrm{d}x} &= \simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+x^{m} *{2*n*a}*x* ({a} * x^2+{b})^{n-1}}\\
&= \simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+{2*n*a}*x^{m+1}* ({a} * x^2+{b})^{n-1}}\\
&= \simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m}*({a}*x^2+{b})+{2*n*a}x^{2})} \\
&= \simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m*a+2*a*n}*x^2+{m*b})}
\end{align}

Hence $\simplify[std]{g(x)={m*a+2*a*n}*x^2+{m*b}}$

Question 2

Differentiate the following function $f(x)$ using the product rule.

$\simplify[dPoly]{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$

You are given that \[\simplify[dPoly]{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}\]

for a polynomial $g(x)$. You have to find $g(x)$.

$g(x)=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{4 - 8 x}$

Clicking on Show steps gives you more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[dPoly]{u = ({a} + {b} * x) ^ {m}}\Rightarrow \simplify[dPoly]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\]

\[\simplify{v = e ^ ({n} * x)} \Rightarrow \simplify{Diff(v,x,1) = {n} * e ^ ({n} * x)}\]

Hence on substituting into the product rule above we get:

\[\simplify[dPoly]{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x) = ({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}\]

The last step was to take out the common term $\simplify[dPoly]{({a} + {b} * x) ^ {m -1} * e ^ ({n} * x)}$.

Hence \[\simplify[dPoly]{g(x) = {m * b + n * a} + {n * b} * x}\].

Question 3

Differentiate the following function $f(x)$ using the product rule.

$\simplify[std]{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$

$\displaystyle \frac{df}{dx}=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{5 \left ( 5 - 4 x \right )^{ 8 } e^{ 5 x } - 32 \left ( 5 - 4 x \right )^{ 7 } e^{ 5 x }}$

Clicking on Show steps gives you more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[std]{u = ({a} + {b} * x) ^ {m}}\Rightarrow \simplify[std]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\]

\[\simplify[std]{v = e ^ ({n} * x)} \Rightarrow \simplify[std]{Diff(v,x,1) = {n} * e ^ ({n} * x)}\]

Hence on substituting into the product rule above we get:

\[\simplify[std]{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x) }\]

Question 4

Differentiate the following function $f(x)$ using the product rule.

$\simplify[std]{f(x) = x^{m}sin({b}x + {a}) * e ^ ({n} * x)}$

The answer is of the form:

$\displaystyle \frac{df}{dx}= \simplify[std]{x^{m-1}e^({n}x)g(x)}$ for a function $g(x)$. You have to find $g(x)$

$g(x)=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{\left ( 4 x + 5 \right ) \sin \left ( 9 - 3 x \right ) - 3 x \cos \left ( 9 - 3 x \right )}$

if you input a function of the form $xf(x)$ where $f(x)$ is a function, then you must input it as $x*f(x)$ with * for multiplication e.g. input $x*\sin(ax+b)$ and not $xsin(ax+b)$.

Click on Show steps for more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:
\[\simplify[std]{u = x^{m}} \Rightarrow \simplify[std]{Diff(u,x,1) = {m}x^{m-1}}\]

\[\simplify[std]{v = sin({b} * x+{a})e^({n}x)}\Rightarrow \simplify[std]{Diff(v,x,1) = {b} * cos({b} * x+{a})e^({n}x)+{n}sin({b}x+{a})e^({n}x)}\]

Hence on substituting into the product rule above we get:

\[\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{{m}x^{m-1}sin({b} * x+{a})e^({n}x)+x^{m}({b} * cos({b} * x+{a}) * e ^ ({n} * x) + {n} * sin({b} * x+{a}) * e ^ ({n} * x))}\\ &=&\simplify[std]{x^{m-1}e^({n}x)({b}x*cos({b}x+{a})+{n}x*sin({b}x+{a})+{m}sin({b}x+{a}))}\\ &=&\simplify[std]{x^{m-1}e^({n}x)({b}x*cos({b}x+{a})+({n}x+{m})*sin({b}x+{a}))} \end{eqnarray*}\]
Hence $g(x)=\simplify[std]{{b}x*cos({b}x+{a})+({n}x+{m})*sin({b}x+{a})}$

Question 5

Differentiate the following function $f(x)$ using the product rule.

$\simplify[std]{f(x) = ({a} + {b} * x) ^ {m} * sin({n} * x)}$

$\displaystyle \frac{df}{dx}=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{3 \left ( 4 + x \right )^{ 2 } \sin \left ( 3 x \right ) + 3 \left ( 4 + x \right )^{ 3 } \cos \left ( 3 x \right )}$

Clicking on Show steps gives you more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[std]{u = ({a} + {b} * x) ^ {m}}\Rightarrow \simplify[std]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\]

\[\simplify[std]{v = sin({n} * x)} \Rightarrow \simplify[std]{Diff(v,x,1) = {n} * cos({n} * x)}\]

Hence on substituting into the product rule above we get:

\[\simplify[std]{Diff(f,x,1) = {m*b}({a} + {b} * x) ^ {m-1} * sin({n} * x)+{n}*({a} + {b} * x) ^ {m} * cos({n} * x)}\]

Question 6

Write down the derivatives of the following functions $f(x)$ .

Note that in order to input the square of a function such as $\sinh(x)$ you have to input it as $(\sinh(x))^2$, similarly for the other hyperbolic functions.

a)

$f(x)=\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}$

$\displaystyle{\frac{df}{dx}=\;\;}$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{6 x^{ 5 } \sinh \left ( -3 x + 3 \right ) - 3 x^{ 6 } \cosh \left ( -3 x + 3 \right )}$

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b)

$f(x)=\tanh(\simplify[std]{{a}x+{b}})$

$\displaystyle{\frac{df}{dx}=\;\;}$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{6 \operatorname{sech}^{2}\left( 6 x - 6 \right)}$

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c)

$f(x)=\ln(\cosh(\simplify[std]{{a2}x+{b2}}))$

$\displaystyle{\frac{df}{dx}=\;\;}$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{2 \tanh \left ( 2 x - 9 \right )}$

Score: 0/5

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Advice

Here is a table of the derivatives of some of the hyperbolic functions:

$f(x)$	$\displaystyle{\frac{df}{dx}}$
$\sinh(bx)$	$b\cosh(bx)$
$\cosh(bx)$	$b\sinh(bx)$
$\tanh(bx)$	$\simplify{b*sech(bx)^2}$

a) $\displaystyle{f(x)=\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}}$

Using the product rule we obtain:

\[\frac{df}{dx} = \simplify[std]{{n} * (x ^ {(n -1)}) * sinh({a1} * x + {b1}) + {a1} * (x ^ {n}) * cosh({a1} * x + {b1})}\]

b) $\displaystyle{f(x)=\tanh(\simplify[std]{{a}x+{b}})}$

Using the table we can immediately write the derivative as:

\[\frac{df}{dx} = \simplify[std]{{a}*sech({a}x+{b})^2}\]

c) $\displaystyle{f(x)=\ln(\cosh(\simplify[std]{{a2}x+{b2}})}$

Here we employ the chain rule. We set $u=\cosh(\simplify[std]{{a2}x+{b2}})$, such that $f(x)=f(u)=\ln u$. Then, according to the chain rule:

\[\frac{df}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \]

Evaluating the derivatives on the right-hand side: $\displaystyle{\frac{du}{dx}=\simplify[std]{{a2}*sinh({a2}x+{b2})}}$ and $\displaystyle{\frac{df}{du}=\frac{1}{u}=\frac{1}{\cosh(\simplify[std]{{a2}x+{b2}})}}$.

Then, inserting these into the chain rule gives:

\[\begin{eqnarray*}\frac{df}{dx} &=& \frac{\simplify[std]{{a2} * sinh({a2} * x + {b2})}}{\cosh(\simplify[std]{{a2}x+{b2}})} \\ &=& \simplify[std]{{a2} * tanh({a2} * x + {b2})}\end{eqnarray*}\]

Question 7

Write down the derivatives of the following functions $f(x)$ .

Note that in order to input the square of a function such as $\sinh(x)$ you have to input it as $(\sinh(x))^2$, similarly for the other hyperbolic functions.

a)

$f(x)=\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}$

$\displaystyle{\frac{df}{dx}=\;\;}$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{4 x^{ 3 } \sinh \left ( -9 x + 8 \right ) - 9 x^{ 4 } \cosh \left ( -9 x + 8 \right )}$

Score: 0/1

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b)

$f(x)=\tanh(\simplify[std]{{a}x+{b}})$

$\displaystyle{\frac{df}{dx}=\;\;}$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{5 \operatorname{sech}^{2}\left( 5 x + 7 \right)}$

Score: 0/1

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c)

$f(x)=\ln(\cosh(\simplify[std]{{a2}x+{b2}}))$

$\displaystyle{\frac{df}{dx}=\;\;}$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{4 \tanh \left ( 4 x + 9 \right )}$

Score: 0/1

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Advice

Here is a table of the derivatives of some of the hyperbolic functions:

$f(x)$	$\displaystyle{\frac{df}{dx}}$
$\sinh(bx)$	$b\cosh(bx)$
$\cosh(bx)$	$b\sinh(bx)$
$\tanh(bx)$	$\simplify{b*sech(bx)^2}$

a)

$f(x)=\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}$

Use the product rule to obtain:
\[\frac{df}{dx} = \simplify[std]{{n} * (x ^ {(n -1)}) * sinh({a1} * x + {b1}) + {a1} * (x ^ {n}) * Cosh({a1} * x + {b1})}\]

b)

$f(x)=\tanh(\simplify[std]{{a}x+{b}})$

Using the table above we get:
\[\frac{df}{dx} = \simplify[std]{{a}*sech({a}x+{b})^2}\]

c)

$f(x)=\ln(\cosh(\simplify[std]{{a2}x+{b2}}))$

Using the chain rule we find:

\[\frac{df}{dx} = \simplify[std]{{a2} * tanh({a2} * x + {b2})}\]

Question 8

Differentiate the following function $f(x)$ using the product rule.

$\displaystyle \simplify[std]{f(x) = x ^ {m} * ({a} * x+{b})^{n}}$

$\displaystyle \frac{df}{dx}=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{7 x^{ 6 } \left ( 8 x + 6 \right )^{ 5 } + 40 x^{ 7 } \left ( 8 x + 6 \right )^{ 4 }}$

Clicking on Show steps gives you more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[std]{u = x ^ {m}}\Rightarrow \simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\]

\[\simplify[std]{v = ({a} * x+{b})^{n}} \Rightarrow \simplify[std]{Diff(v,x,1) = {n*a} * ({a} * x+{b})^{n-1}}\]

Hence on substituting into the product rule above we get:

\[\simplify[std]{Diff(f,x,1) = {m}x ^ {m-1} * ({a} * x+{b})^{n}+{n*a}x^{m} * ({a} * x+{b})^{n-1}}\]

Question 9

Differentiate the following function $f(x)$ using the product rule.

$\simplify[std]{f(x) = x ^ {m} * cos({a} * x+{b})}$

$\displaystyle \frac{df}{dx}=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{3 x^{ 2 } \cos \left ( 6 x + 9 \right ) - 6 x^{ 3 } \sin \left ( 6 x + 9 \right )}$

Clicking on Show steps gives you more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[std]{u = x ^ {m}}\Rightarrow \simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\]

\[\simplify[std]{v = cos({a} * x+{b})} \Rightarrow \simplify[std]{Diff(v,x,1) = -{a} * sin({a} * x+{b})}\]

Hence on substituting into the product rule above we get:

\[\simplify[std]{Diff(f,x,1) = {m}x ^ {m-1} * cos({a} * x+{b})-{a}x^{m} * sin({a} * x+{b})}\]

Question 10

Differentiate the following function $f(x)$ using the product rule.

$\simplify[std]{f(x) = sin({b}x + {a}) * e ^ ({n} * x)}$

$\displaystyle \frac{df}{dx}=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{4 \cos \left ( 5 + 4 x \right ) e^{ -5 x } - 5 \sin \left ( 5 + 4 x \right ) e^{ -5 x }}$

Click on Show steps for more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[std]{u = sin({a} + {b} * x)}\Rightarrow \simplify[std]{Diff(u,x,1) = {b} * cos({a} + {b} * x)}\]

\[\simplify[std]{v = e ^ ({n} * x)} \Rightarrow \simplify[std]{Diff(v,x,1) = {n} * e ^ ({n} * x)}\]

Hence on substituting into the product rule above we get:

\[\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{{b} * cos({a} + {b} * x) * e ^ ({n} * x) + {n} * sin({a} + {b} * x) * e ^ ({n} * x)}\\ &=&\simplify[std]{({b}cos({a}+{b}x)+{n}sin({a}+{b}x))e^({n}x)} \end{eqnarray*}\]

Question 11

Differentiate the following function $f(x)$ using the product rule.

$\simplify[dPoly]{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$

You are given that \[\simplify[dPoly]{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}\]

for a polynomial $g(x)$. You have to find $g(x)$.

$g(x)=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-14 - 24 x}$

Click on Show steps to get more information, you will not lose any marks by doing so.

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

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Advice

The product rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\]

For this example:

\[\simplify[dPoly]{u = ({a} + {b} * x) ^ {m}}\Rightarrow \simplify[dPoly]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\]

\[\simplify{v = e ^ ({n} * x)} \Rightarrow \simplify{Diff(v,x,1) = {n} * e ^ ({n} * x)}\]

Hence on substituting into the product rule above we get:

\[\simplify[dPoly]{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x) = ({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}\]

The last step was to take out the common term $\simplify[dPoly]{({a} + {b} * x) ^ {m -1} * e ^ ({n} * x)}$.

Hence \[\simplify[dPoly]{g(x) = {m * b + n * a} + {n * b} * x}\].

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Maria's copy of Differentiation using the product rule

Maria's copy of Differentiation using the product rule

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