Find me midpoint of -9 and 6
2 answers:
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Answer:
a) d(sinh(f(x)))/dx = cosh(f(x))·df(x)/dx
b) d(cosh(f(x))/dx = sinh(f(x))·df(x)/dx
c) d(tanh(f(x))/dx = sech(f(x))²·df(x)/dx
d) d(sech(4x+2))/dx = -4sech(4x+2)tanh(4x+2)
Step-by-step explanation:
To do these, you need to be familiar with the derivatives of hyperbolic functions and with the chain rule.
The chain rule tells you that ...
(f(g(x)))' = f'(g(x))g'(x) . . . . where the prime indicates the derivative
The attached table tells you the derivatives of the hyperbolic trig functions, so you can answer the first three easily.
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a) sinh(u)' = sinh'(u)·u' = cosh(u)·u'
For u = f(x), this becomes ...
sinh(f(x))' = cosh(f(x))·f'(x)
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b) After the same pattern as in (a), ...
cosh(f(x))' = sinh(f(x))·f'(x)
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c) Similarly, ...
tanh(f(x))' = sech(f(x))²·f'(x)
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d) For this one, we need the derivative of sech(x) = 1/cosh(x). The power rule applies, so we have ...
sech(x)' = (cosh(x)^-1)' = -1/cosh(x)²·cosh'(x) = -sinh(x)/cosh(x)²
sech(x)' = -sech(x)·tanh(x) . . . . . basic formula
Now, we will use this as above.
sech(4x+2)' = -sech(4x+2)·tanh(4x+2)·(4x+2)'
sech(4x+2)' = -4·sech(4x+2)·tanh(4x+2)
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Here we have used the "prime" notation rather than d( )/dx to indicate the derivative with respect to x. You need to use the notation expected by your grader.
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<em>Additional comment on notation</em>
Some places we have used fun(x)' and others we have used fun'(x). These are essentially interchangeable when the argument is x. When the argument is some function of x, we mean fun(u)' to be the derivative of the function after it has been evaluated with u as an argument. We mean fun'(u) to be the derivative of the function, which is then evaluated with u as an argument. This distinction makes it possible to write the chain rule as ...
f(u)' = f'(u)u'
without getting involved in infinite recursion.
