1. y = x^x
2. y = (sin^2xtan^4x)/(x^2+1)^2
3. y = x^(1/x)
4. y = 2^(-x^3)
Okay, so I'm pretty sure 1 is done easily:
ln y = ln (x^x)
ln y = x ln x
d/dx ln y = d/dx x ln x
y'(1/y) = x(1/x) + ln x(1) = 1 + ln x
y'(1/y) = ln x + 1
y' y(1/y) = (ln x + 1)y
y' = (ln x + 1)y
since y = x^x,
y' = (ln x + 1)x^x
For #2, this is as far as I got:
ln y = ln[(sin^2xtan^4x)/(x^2+1)^2
y'(1/y) = ln(sin^2xtan^4x) - ln(x^2+1)^2
What do I do after that?
#3, I thought was like #1, but I still think my answer is incorrect or maybe I didn't simplify properly:
ln y = (1/x) ln x
d/dx ln y = d/dx (1/x) ln x
y'(1/y) = (1/x)*(1/x) + ln x*(-1/x^2) = (1/x)^2 + ln x*(-1/x^2)
y' = (1/x)^2 + ln x*(-1/x^2)y
y' = (1/x)^2 + ln x*(-1/x^2)x^(1/x)
And for #4 I am completely lost since there is an exponent within the exponent. Any kind of hints or clues would be SO appreciated.
#1 is good
in #2 you could go further before taking any derivative
y'(1/y) = ln(sin^2xtan^4x) - ln(x^2+1)^2 or
y'(1/y) = ln(sin^2x) + ln(tan^4x) - 2ln(x^2 + 1)
y'(1/y) = 2ln(sinx) + 4ln(tanx) - 2ln(x^2+1)
and now differentiate it
In #3 from your second last line of
y' = (1/x)^2 + ln x*(-1/x^2)y I would do
y ' = y(1 - lnx)/x^2 and leave it like that
#4 is exponential differentiation.
Notice the base is a constant, not a variable
recall if y = a^u
then y' = (ln a)(a^u)(u')
so for y = 2^(-x^3)
y' = ln2(-3x^2)(2^(-x^3))
= -3ln2(x^2)(2^(-x^3))
If you want to do it by the same method as the above, you would get
lny = ln(2^(-x^3))
= (-x^3)(ln2)
= -ln2(x^3)
now take the derivative
y'/y = -ln2(3)x^2
y' = -3ln2(x^2)y
= my answer above
For #2:
To differentiate the expression, you can use the quotient rule for differentiation. The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), then the derivative of f(x) with respect to x is given by:
f'(x) = (g'(x)h(x) - g(x)h'(x))/[h(x)]^2
In this case, g(x) = sin^2(x)tan^4(x) and h(x) = (x^2 + 1)^2.
Let's calculate the derivatives step-by-step:
g'(x) = 2sin(x)cos(x)tan^4(x) + sin^2(x)4tan^3(x)sec^2(x)
h'(x) = 2(x^2 + 1)(2x)
Now, substitute these values into the quotient rule formula:
f'(x) = [(2sin(x)cos(x)tan^4(x) + sin^2(x)4tan^3(x)sec^2(x))(x^2 + 1)^2 - sin^2(x)tan^4(x)(2(x^2 + 1)(2x))]/[(x^2 + 1)^2]^2
Simplifying this expression will give you the final derivative.
For #3:
To differentiate y = x^(1/x), you can use the chain rule for differentiation. The chain rule states that if you have a function of the form f(g(x)), the derivative of f(g(x)) with respect to x is given by:
[f(g(x))]' = f'(g(x)) * g'(x)
In this case, f(u) = u^(1/u) and g(x) = x.
Let's calculate the derivatives step-by-step:
f'(u) = (1/u) * u^(1/u - 1) - ln(u) * u^(1/u) / u^2
= u^(-1/u) - ln(u) * u^(1/u) / u^2
g'(x) = 1
Now, substitute these values into the chain rule formula:
[y]' = (u^(-1/u) - ln(u) * u^(1/u) / u^2) * 1
Since u = x, we can substitute x for u:
[y]' = (x^(-1/x) - ln(x) * x^(1/x) / x^2)
This is the final derivative.
For #4:
To differentiate y = 2^(-x^3), you can use the chain rule for differentiation. The chain rule states that if you have a function of the form f(g(x)), the derivative of f(g(x)) with respect to x is given by:
[f(g(x))]' = f'(g(x)) * g'(x)
In this case, f(u) = 2^u and g(x) = -x^3.
Let's calculate the derivatives step-by-step:
f'(u) = ln(2) * 2^u
g'(x) = -3x^2
Now, substitute these values into the chain rule formula:
[y]' = (ln(2) * 2^(-x^3)) * (-3x^2)
This is the final derivative.
For #2, you are on the right track. After you took the natural logarithm of both sides, you need to differentiate the equation implicitly with respect to x to find y'.
ln y = ln[(sin^2x * tan^4x)/(x^2+1)^2]
Differentiating implicitly:
(d/dx) ln y = (d/dx) ln[(sin^2x * tan^4x)/(x^2+1)^2]
Using the chain rule:
y' * (1/y) = [(2sinx * cosx * tan^3x)/(x^2+1)^2] - [2x/(x^2+1)]
Rearranging the equation to solve for y':
y' = y * [(2sinx * cosx * tan^3x)/(x^2+1)^2 - (2x/(x^2+1))]
For #3, again, you are on the right track. After taking the natural logarithm of both sides, differentiate the equation implicitly with respect to x to find y'.
ln y = (1/x) ln x
Differentiating implicitly:
(d/dx) ln y = (d/dx) ((1/x) ln x)
Using the product rule and chain rule:
y' * (1/y) = [(1/x) * (1/x) + (1/x) * (d/dx) ln x]
Simplifying:
y' = y * [(1/x^2) + (1/x^2) * (1/x)]
y' = y * [(1/x^2) + (1/x^3)]
For #4, you can use the chain rule and power rule to differentiate the equation.
y = 2^(-x^3)
Taking the natural logarithm of both sides:
ln y = ln (2^(-x^3))
Using the property of logarithms, ln(y) = -x^3 * ln(2)
Differentiating implicitly:
(d/dx) ln y = (d/dx) (-x^3 * ln(2))
Using the chain rule and power rule:
y' * (1/y) = -3x^2 * ln(2)
Simplifying:
y' = y * (-3x^2 * ln(2))
y' = 2^(-x^3) * (-3x^2 * ln(2))
These are the derivatives for each of the given equations.