Maths Support: Diagnostic test - Differentiation

Question 1

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)

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

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b)

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

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c)

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

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

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

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

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

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

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

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

Using the chain rule we find:

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

Question 2

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

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

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$\simplify[std]{f(x) = ln(({a}x+{b})^{m})}$
First note that we can simplify this by using the rule that $\simplify[std]{ln(a^r)=r*ln(a)}$.
Hence $\simplify[std]{f(x) = ln(({a}x+{b})^{m})={m}ln({a}x+{b})}$
So we need to differentiate $\simplify[std]{ln({a}x+{b})}$
The chain rule says that if $f(x)=g(h(x))$ then
\[\simplify[std]{f'(x) = h'(x)g'(h(x))}\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\[\frac{df}{dx} = \frac{du}{dx}\frac{df(u)}{du}\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

For this example, we let $u=\simplify[std]{{a}x +{b}}$ and we have $f(u)=\simplify[std]{{m}*ln(u)}$.
This gives
\[\begin{eqnarray*}\frac{du}{dx} &=& \simplify[std]{{a}}\\ \frac{df(u)}{du} &=& \simplify[std]{{m}/u} \end{eqnarray*}\]

Hence on substituting into the chain rule above we get:

\[\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{({a}) * ({m}/u)}\\ &=& \simplify[std]{{a*m}/({a}x+{b})} \end{eqnarray*}\]
on replacing $u$ by $\simplify[std]{{a}x+{b}}$.

Question 3

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

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

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Advice

$\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}$
The chain rule says that if $f(x)=g(h(x))$ then
\[\simplify[std]{f'(x) = h'(x)g'(h(x))}\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\[\frac{df}{dx} = \frac{du}{dx}\frac{df(u)}{du}\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

For this example, we let $u=\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\simplify[std]{sqrt(u)=u^{1/2}}$.
This gives
\[\begin{eqnarray*}\frac{du}{dx} &=& \simplify[std]{{m*a}x ^ {m -1}}\\ \frac{df(u)}{du} &=& \simplify[std]{{1/2}*u^{-1/2}=1/(2*sqrt(u))} \end{eqnarray*}\]

Hence on substituting into the chain rule above we get:

\[\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{{m*a}x ^ {m-1} * (1/(2*sqrt(u)))}\\ &=&\simplify[std]{{m*a}x^{m-1}/(2*sqrt(u))}\\ &=& \simplify[std]{({a*m}x ^ {m-1})/(2*sqrt({a} * x^{m}+{b}))} \end{eqnarray*}\]
on replacing $u$ by $\simplify[std]{{a}x^{m}+{b}}$.

Question 4

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

\[\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}\]

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Advice

$\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$
The chain rule says that if $f(x)=g(h(x))$ then
\[\simplify[std]{f'(x) = h'(x)g'(h(x))}\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\[\frac{df}{dx} = \frac{du}{dx}\frac{df(u)}{du}\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

For this example, we let $u=\simplify[std]{{a}x^{m} +{b}x^2+{c}}$ and we have $f(u)=\simplify[std]{e^u}$.
This gives
\[\begin{eqnarray*}\frac{du}{dx} &=& \simplify[std]{{a*m}x^{m-1} +{2*b}x}\\ \frac{df(u)}{du} &=& \simplify[std]{e^u} \end{eqnarray*}\]

Hence on substituting into the chain rule above we get:

\[\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{({a*m}x^{m-1} +{2*b}x) * (e^u)}\\ &=& \simplify[std]{({a*m}x^{m-1} +{2*b}x)*e^({a}x^{m} +{b}x^2+{c})} \end{eqnarray*}\]
on replacing $u$ by $\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.

Question 5

Differentiate the following function $f(x)$.

$\simplify[std]{f(x) = x ^ {m} * ({a} * x^2+{b})^{n}}$
The answer is in the form
\[\frac{df}{dx}=\simplify[std]{x^{m-1}({a}x^2+{b})^{n-1}*g(x)}\] for a polynomial $g(x)$.

You have to find $g(x)$.

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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:

\[\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+x^{m} *{2*n*a}*x* ({a} * x^2+{b})^{n-1}}\\ &=& \simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+{2*n*a}*x^{m+1}* ({a} * x^2+{b})^{n-1}}\\ &=& \simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m}*({a}*x^2+{b})+{2*n*a}x^{2})} \\ &=&\simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m*a+2*a*n}*x^2+{m*b})} \end{eqnarray*}\]

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

Question 6

Differentiate the following function $f(x)$ using the product rule. The answer will be of the form
\[\simplify[std]{x^{m-1}({a}x+{b})^{n-1}g(x)}\] for a polynomial $g(x)$. You have to find $g(x)$

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

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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:

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

Question 7

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

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

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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 8

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

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

The answer is in the form \[\frac{df}{dx}=\simplify[std]{ x^{m-1}/(2*sqrt({a}x+{b}))g(x)}\]
for a polynomial $g(x)$. You have to find $g(x)$.

$g(x)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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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:

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

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

Hence on substituting into the product rule above we get:

\[\begin{eqnarray*} \frac{df}{dx}&=& \simplify[std]{{m}x ^ {m-1} * sqrt({a} * x+{b})+{a/2}x^{m} * ({a} * x+{b})^{-1/2}}\\ &=& \simplify[std]{{m}x ^ {m-1} * sqrt({a} * x+{b})+{a}x^{m}/(2*sqrt({a} * x+{b}))}\\ &=& \simplify[std]{(2*{m}x^{m-1}({a}x+{b})+ {a}x^{m})/(2*sqrt({a} * x+{b}))}\\ &=&\simplify[std]{(x^{m-1}({2*m}({a}x+{b})+{a}x))/(2*sqrt({a} * x+{b}))}\\ &=&\simplify[std]{x^{m-1}/(2*sqrt({a} * x+{b}))({2*m*a+a}x+{2*m*b})} \end{eqnarray*}\]
Hence $g(x)=\simplify[std]{{2*m*a+a}x+{2*m*b}}$

Question 9

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

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

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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 10

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

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

$\displaystyle \frac{df}{dx}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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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 11

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

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

$\displaystyle \frac{df}{dx}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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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 12

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

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

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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 13

Differentiate the following function $f(x)$ using the product rule. The answer will be of the form
\[\simplify[std]{({a}x+{b})^{m-1}({c}x+{d})^{n-1}g(x)}\] for a polynomial $g(x)$. You have to find $g(x)$

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

$g(x)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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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:

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

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

Hence on substituting into the product rule above we get:

\[\simplify[std]{Diff(f,x,1) = {m*a}({a}x+{b}) ^ {m-1} * ({c} * x+{d})^{n}+{n*c}({a}x+{b})^{m} * ({c} * x+{d})^{n-1}=({a}x+{b})^{m-1}({c}x+{d})^{n-1}({m*a*c+n*c*a}x+{m*a*d+n*c*b})}\]
on taking out the common terms $\simplify[std]{({a}x+{b})^{m-1}({c}x+{d})^{n-1}}$.
So $\simplify[std]{g(x)= {m*a*c+n*c*a}x+{m*a*d+n*c*b}}$

Question 14

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 $g(x)$.

$g(x)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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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 15

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

\[\simplify[std]{f(x) = ({a} * x+{b})/({c}x^2+{d}x+{f})}\]
You are given that \[\frac{df}{dx}=\simplify[std]{g(x)/({c}x^2+{d}x+{f})^2}\]
for a polynomial $g(x)$. You are asked to find $g(x)$

$g(x)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Input numbers as fractions or integers and not as decimals.

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The quotient rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\]

For this example:

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

\[\simplify[std]{v = ({c} * x^2+{d}x+{f})} \Rightarrow \simplify[std]{Diff(v,x,1) = {2*c}x+{d}}\]

Hence on substituting into the quotient rule above we get:

\[\begin{eqnarray*} \frac{df}{dx}&=&\simplify[std]{({a}({c}x^2+{d}x+{f})-({2*c}x+{d})({a}x+{b}))/({c}x^2+{d}x+{f})^2}\\ &=&\simplify[std]{({a*c}x^2+{a*d}x+{a*f}-{2*c*a}x^2-{a*d+2*c*b}x+{d*b})/({c}x^2+{d}x+{f})^2}\\ &=&\simplify[std]{({-c*a}x^2+{-2*b*c}x+{a*f-d*b})/({c}x^2+{d}x+{f})^2} \end{eqnarray*}\]
Hence $g(x)=\simplify[std]{{-c*a}x^2+{-2*b*c}x+{a*f-d*b}}$

Question 16

Differentiate the following function $f(x)$ using the quotient rule or otherwise.

\[\simplify[std]{f(x) = ({a} * x + {b}) / Sqrt({c} * x + {d})}\]

You are given that \[\simplify[std]{Diff(f,x,1) = g(x) / (2 * ({c} * x + {d}) ^ (3 / 2))}\]

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

Input all numbers as fractions or integers.

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$g(x)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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

\[\simplify[std]{Diff(u/v,x,1)=(v * Diff(u,x,1) -(u * Diff(v,x,1))) / v ^ 2}\]

For this example:

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

\[\simplify[std]{v = Sqrt({c} * x + {d})} \Rightarrow \simplify[std]{Diff(v,x,1) = {c} / (2 * Sqrt({c} * x + {d}))}\]

Hence on substituting into the quotient rule above we get:

\[\simplify[std]{Diff(f,x,1) = ({a} * Sqrt({c} * x + {d}) -(({a} * x + {b}) * Diff(v,x,1))) / ({c} * x + {d}) = ({a} * Sqrt({c} * x + {d}) -(({c} * ({a} * x + {b})) / (2 * Sqrt({c} * x + {d})))) / ({c} * x + {d}) = ({2 * a} * ({c} * x + {d}) -({c} * ({a} * x + {b}))) / (2 * ({c} * x + {d}) ^ (3 / 2)) = ({a * c} * x + {2 * a * d -(c * b)}) / (2 * ({c} * x + {d}) ^ (3 / 2))}\]

Hence \[\simplify[std]{g(x) = {a * c} * x + {2 * a * d -(c * b)}}\].

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