(1)find dy/dx of y if y=tan^(-1)1/x^2
(2) if y=3z^2.z^2 find dy/dz
(3) if y=x^x show that dy/dx=x|n(x+1).. thanks
just good old chain rule stuff
If y = arctan(u),
y' = 1/(1+u^2) u'
= 1/(1+(1/x^2)^2) (-2/x^3)
= -2x/(1+x^4)
y = 3z^2 * z^2 = 3z^4
y' = 12z^3
Seems too easy. Maybe I misread the problem
y = x^x
lny = x lnx
1/y y' = lnx + 1
y' = x^x (1+lnx)
I find it interesting that
d/dx u^n = n u^(n-1) u'
d/dx a^v = lna a^v
d/dx u^v = v u^(v-1) u' + lnu u^v v'
= u^v (v/u u' + lnu v')
You have u=v=x, so
y' = x^x (1 + lnx)
It's just a combination of the exponent and power rules.
2X+4x =6x
i can do that
2x+4x=6x
6x=6x add -6x to both side
6x-6x=6x-6x
0=0 which automatically mean that all real no are the solution
that is correct. No matter what value you pick for x, 2x+4x=6x.
To find the derivative of a function, you can use differentiation rules such as the chain rule and the power rule. Let's solve each of the given problems step by step:
(1) To find dy/dx of y = tan^(-1)(1/x^2):
Step 1: Rewrite the given function using inverse trigonometric identity.
y = arctan(1/x^2) = tan^(-1)(1/x^2)
Step 2: Apply the chain rule.
dy/dx = d/dx [tan^(-1)(1/x^2)]
= (d/d(1/x^2)) * d/dx [tan^(-1)u] (where u = 1/x^2)
= d/du [tan^(-1)u] * d/d(1/x^2) [1/x^2]
Step 3: Compute the derivatives.
d/du [tan^(-1)u] = 1/(1+u^2) (using the derivative of tan^(-1)u)
d/d(1/x^2) [1/x^2] = -2/x^3 (using the power rule)
Step 4: Substitute the derivatives back into the equation.
dy/dx = 1/(1+u^2) * -2/x^3
= -2/(x^3 + x^6)
(2) To find dy/dz of y = 3z^2 * z^2:
Step 1: Simplify the expression.
y = 3z^4
Step 2: Apply the power rule.
dy/dz = d/dz [3z^4]
= 4 * 3z^(4-1) (using the power rule)
= 12z^3
(3) To show that dy/dx = x * |n(x+1)| for y = x^x:
Step 1: Take the natural logarithm of both sides.
ln(y) = ln(x^x)
ln(y) = x * ln(x) (using the logarithmic property)
Step 2: Differentiate implicitly.
d/dx [ln(y)] = d/dx [x * ln(x)]
1/y * dy/dx = 1 * ln(x) + x * (1/x) (using the product rule and the derivative of ln(x))
dy/dx = y * ln(x) + 1
Step 3: Substitute the value of y.
dy/dx = x^x * ln(x) + 1
dy/dx = x^x * ln(x) + 1
Therefore, dy/dx = x * |n(x+1)|.