Chapter 3: Q34E (page 189)
To find the derivative of the function \(y = sinh(coshx)\).
Short Answer
The derivative of the function \(y = \sinh (\cosh x)\) is \(\sinh x\cosh (\cosh x)\).
Chapter 3: Q34E (page 189)
To find the derivative of the function \(y = sinh(coshx)\).
The derivative of the function \(y = \sinh (\cosh x)\) is \(\sinh x\cosh (\cosh x)\).
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Get started for free(A) to determine the function\(f(x) = \frac{1}{{x - 1}},\;x > 1\)is one-to-one.\({f^{ - 1}}(x) = \frac{1}{x} + 1\)in
(B) To determine the value of\({\left( {{f^{ - 1}}} \right)^\prime }(2)\), where\(f(x) = \frac{1}{{x - 1}}\).
(C) To determine the inverse of the function\(f(x) = 9 - {x^2}\)and state its domain and range.
(D) To determine whether the value of\({\left( {{f^{ - 1}}} \right)^\prime }(2)\)is\(\frac{1}{4}\)using the inverse function.
(E) To sketch: The graph of\(f(x) = \frac{1}{{x - 1}}\)and\({f^{ - 1}}(x) = \frac{1}{x} + 1\)in the same coordinate.
1โ38 โ Find the limit. Use lโHospitalโs Rule where appropriate. If there is a more elementary method, consider using it. If lโHospitalโs Rule doesnโt apply, explain why.
7.\(\mathop {lim}\limits_{\theta \to \frac{\pi }{2}} \frac{{1 - sin\theta }}{{1 + cos2\theta }}\).
Determine the inverse of a function \(m = f(v) = \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\) and interpret the meaning.
To determine the value of \(\mathop {\lim }\limits_{u \to \infty } \frac{{\left( {\frac{{{\varepsilon ^u}}}{{10}}} \right)}}{{{u^3}}}\).
(a) To determine the graph of functions \(cschx, sech x\), and \(\coth x\) by using graph of the functions \(\sinh x,\cosh x\), and \(\tanh x\).
(b) To check the graphs sketched in part (a) and using graphing device to draw the graphs.
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