1.) Find the derivative of tan (sec x).
2.) Find the derivative if 1/x in four ways, using the limit process, power rule, quotient rule and implicit differentiation.
3.) Show that the derivative of sec^-1 x is 1/(|x|*sqrt(x^2 -1)).
4. Find the derivative of 2^(e^(sin x)).
Thank you in advance.
d/du(tan u) = sec^2 u
so, use that and apply the chain rule:
sec^2(sec x) * secx tanx
I'm sure you can do #2
y = sec^-1 x
sec y = x
secy tany y' = 1
y' = 1/(secy tany)
y' = 1/(x √(x^2-1))
y = 2^(e^(sinx))
y' = ln2 2^(e^(sinx)) * e^(sinx) * cosx
1.) To find the derivative of tan(sec(x)), we can use the chain rule. The chain rule states that if we have a composition of functions, f(g(x)), then the derivative is given by f'(g(x)) * g'(x).
Let's break down the function tan(sec(x)) as follows:
f(u) = tan(u)
g(x) = sec(x)
Now, we need to find the derivatives of f(u) and g(x):
f'(u) = sec^2(u)
g'(x) = sec(x) * tan(x)
Using the chain rule, the derivative of the composition is:
[tan(sec(x))]'
= f'(g(x)) * g'(x)
= sec^2(sec(x)) * sec(x) * tan(x)
Thus, the derivative of tan(sec(x)) is sec^2(sec(x)) * sec(x) * tan(x).
2.) Let's find the derivative of 1/x in four different ways:
a) Using the limit process:
To find the derivative of 1/x using the limit process, recall that the derivative of a function f(x) is given by:
f'(x) = lim(h->0) [f(x+h) - f(x)] / h
For 1/x, we have:
f(x) = 1/x
f(x+h) = 1/(x+h)
Now, let's substitute these values into the limit definition:
f'(x) = lim(h->0) [1/(x+h) - 1/x] / h
Simplifying this expression, we get:
f'(x) = lim(h->0) [x - (x+h)] / (x(x+h)h)
= lim(h->0) [-h/(x(x+h)h)]
= lim(h->0) [-1/(x(x+h))]
= -1/(x^2)
Therefore, the derivative of 1/x using the limit process is -1/(x^2).
b) Using the power rule:
The power rule states that if we have a function f(x) = x^n, then the derivative is given by f'(x) = nx^(n-1).
For 1/x, we can rewrite it as x^(-1), where n = -1.
f'(x) = (-1)x^(-1-1)
= -x^(-2)
= -1/x^2
So, the derivative of 1/x using the power rule is -1/x^2.
c) Using the quotient rule:
The quotient rule states that if we have a function f(x) = g(x)/h(x), then the derivative is given by f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / [h(x)]^2.
For 1/x, we can rewrite it as g(x) = 1 and h(x) = x.
g'(x) = 0 (since the derivative of a constant, 1, is zero)
h'(x) = 1
Now, let's substitute these values into the quotient rule formula:
f'(x) = (0 * x - 1 * 1) / x^2
= -1/x^2
Therefore, the derivative of 1/x using the quotient rule is -1/x^2.
d) Using implicit differentiation:
To find the derivative of 1/x using implicit differentiation, we can treat y = 1/x as an implicit function of x.
Differentiating both sides with respect to x:
d/dx (y) = d/dx (1/x)
Using the chain rule on the left side, we get:
dy/dx = -1/x^2
Therefore, the derivative of 1/x using implicit differentiation is -1/x^2.
3.) To show that the derivative of sec^(-1)(x) is 1/(|x| * sqrt(x^2 - 1)), we can use the chain rule.
The function sec^(-1)(x) can also be written as arccos(1/x). Let u = 1/x.
Now, let's derive u with respect to x:
du/dx = d/dx (1/x)
= -1/x^2
Using the chain rule, we have:
[d(arccos(u)) / du] * du/dx
Recall that d(arccos(u)) / du = -1/sqrt(1 - u^2).
Therefore, the derivative of sec^(-1)(x) is:
-1/sqrt(1 - (1/x)^2) * (-1/x^2)
= 1/(|x| * sqrt(x^2 - 1))
4. To find the derivative of 2^(e^(sin(x))), we can use the chain rule.
Let's break down the function as follows:
f(u) = 2^u
g(x) = e^(sin(x))
The derivatives of f(u) and g(x) are:
f'(u) = ln(2) * 2^u
g'(x) = cos(x) * e^(sin(x))
Using the chain rule, the derivative of the composition is:
[2^(e^(sin(x)))]'
= f'(g(x)) * g'(x)
= ln(2) * 2^(e^(sin(x))) * cos(x) * e^(sin(x))
Thus, the derivative of 2^(e^(sin(x))) is ln(2) * 2^(e^(sin(x))) * cos(x) * e^(sin(x)).
I hope that clarifies how to find the derivatives of the given functions! Let me know if you need any further explanation.