$f(x)=\\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}$
\n$\\displaystyle{\\frac{df}{dx}=\\;\\;}$[[0]]
\n ", "scripts": {}, "gaps": [{"answer": "{n} * (x ^ {(n -1)}) * sinh({a1} * x + {b1}) + {a1} * (x ^ {n}) * Cosh({a1} * x + {b1})", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "variableReplacementStrategy": "originalfirst", "answersimplification": "std", "variableReplacements": [], "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0}, {"prompt": "\n$f(x)=\\tanh(\\simplify[std]{{a}x+{b}})$
\n$\\displaystyle{\\frac{df}{dx}=\\;\\;}$[[0]]
\n ", "scripts": {}, "gaps": [{"answer": "{a}*sech({a}x+{b})^2", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "variableReplacementStrategy": "originalfirst", "answersimplification": "std", "variableReplacements": [], "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0}, {"prompt": "\n$f(x)=\\ln(\\cosh(\\simplify[std]{{a2}x+{b2}}))$
\n$\\displaystyle{\\frac{df}{dx}=\\;\\;}$[[0]]
\n ", "scripts": {}, "gaps": [{"answer": "{a2} * tanh({a2} * x + {b2})", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "variableReplacementStrategy": "originalfirst", "answersimplification": "std", "variableReplacements": [], "marks": 5, "vsetrangepoints": 5}], "type": "gapfill", "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0}], "statement": "\n \n \nWrite down the derivatives of the following functions $f(x)$ .
\n \n \n \nNote 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.
\n \n \n ", "tags": ["calculus", "Calculus", "chain rule", "checked2015", "cosh", "derivatives of hyperbolic functions", "differential", "differential ", "differentiate", "differentiating hyperbolic functions", "differentiation", "hyperbolic functions", "MAS1601", "mas1601", "product rule", "query", "sinh", "tanh", "tested1"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "29/06/2012:
\nAdded and edited tags.
\n19/07/2012:
\nAdded description.
\nThere is also the problem of inputting functions of the form $xf(x)$ if $n=1$ or $2$ in the first question. So have reset $n$ to between $3$ and $7$. Otherwise would have to have an instruction here (perhaps depending on value of $n$).
\nChecked calculation.
\n23/07/2012:
\nAdded tags.
\nQuestion appears to be working correctly.
\n22/12/2012:(WHF)
\nChecked calculations, OK. Added tested1 tag.
\nIf users factorise the answer to the first question they may be inputting something of the form xf(x). So may be wise to write an instruction to use x*f(x). Raised as a query via the query tag.
\n", "licence": "Creative Commons Attribution 4.0 International", "description": "
Differentiate the following functions: $\\displaystyle x ^ n \\sinh(ax + b),\\;\\tanh(cx+d),\\;\\ln(\\cosh(px+q))$
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "Here is a table of the derivatives of some of the hyperbolic functions:
\n| $f(x)$ | $\\displaystyle{\\frac{df}{dx}}$ |
|---|---|
| $\\sinh(bx)$ | $b\\cosh(bx)$ |
| $\\cosh(bx)$ | $b\\sinh(bx)$ |
| $\\tanh(bx)$ | $\\simplify{b*sech(bx)^2}$ |
a) $\\displaystyle{f(x)=\\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}}$
\nUsing the product rule we obtain:
\n\\[\\frac{df}{dx} = \\simplify[std]{{n} * (x ^ {(n -1)}) * sinh({a1} * x + {b1}) + {a1} * (x ^ {n}) * cosh({a1} * x + {b1})}\\]
\nb) $\\displaystyle{f(x)=\\tanh(\\simplify[std]{{a}x+{b}})}$
\nUsing the table we can immediately write the derivative as:
\n\\[\\frac{df}{dx} = \\simplify[std]{{a}*sech({a}x+{b})^2}\\]
\nc) $\\displaystyle{f(x)=\\ln(\\cosh(\\simplify[std]{{a2}x+{b2}})}$
\nHere 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:
\n\\[\\frac{df}{dx}=\\frac{df}{du}\\cdot \\frac{du}{dx} \\]
\nEvaluating the derivatives on the right-hand side: $\\displaystyle{\\frac{du}{dx}=\\simplify[std]{{a2}*sinh({a2}x+{b2})}}$ and $\\displaystyle{\\frac{df}{du}=\\frac{1}{u}=\\frac{1}{\\cosh(\\simplify[std]{{a2}x+{b2}})}}$.
\nThen, inserting these into the chain rule gives:
\n\\[\\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*}\\]
\n", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}