Find Derivative
Differentiate y = (lnx)²
y' = (2lnx/x)
Explanation: Use the chain rule to differentiate
Differentiate y = ln(lnx)
y' = (1/xlnx)
Explanation:
The derivative of ln(x) is (1/x). Use the chain rule to differentiate.
y = 32x
y' = ln3(32x)2
Explanation:
Use derivative of au rules to differentiate
What is the tangent line equation for y = ln(√5-x2) at x = 2.
1) Differentiate the function
y' = (1/√5-x2)((1/2)(5-x2)-1/2(-2x)
y' = (-2x)/(2(√5-x2)(√5-x2)
y' = -x/(5-x2)
2)Find the slope of the tangent line at x=2
y'(2) = 2/5-4 = 2
3)Plug in x=2 to the original function to find the y for the tangent equation
y = 3 + ln√5-4
y= 3 =ln1
y=3
4) Put all of the information together for the tangent equation
y-3=-2(x-2)
What is the inverse of the natural log function?
f(x) = ln x is f-1(x) = ex.
Differentiate y = x^(3^x)
y = x(3^x)
lny = ln(x3^(x))
lny = 3xln(x)
(dy/dx)(1/y) = 3x(1/x) + ln(x)(3^x)(ln3)
(dy/dx) = y((3x/x) + ln(x)(3x)(ln3))
Answer:
(dy/dx) = x(3^x) ((3x/x) + ln(x)(3x)(ln3))
Explanation:
Take natural log, drop the exponent, and solve for (dy/dx).
Differentiate y = x3x
y = x3x
lny = ln(x3x)
lny= 3xln(x)
(dy/dx)(1/y) = 3((x)(1/x) + lnx)
(dy/dx) = y(3((x)(1/x) + lnx)
Final Answer:
(dy/dx) = x3x(3 + 3lnx)
Explanation:
Take the natural log of both sides, drop the exponent, and solve for (dy/dx). Remember when you are solving that y = x3x
y = 4x^3 - sinx
y' = ln4( 4x^3 - sinx)(3x2 - cosx)
Explanation:
Use derivative of au rules to differentiate
The tangent line to the graph of y = e2-x at the point (1,e) intersects both coordinate axes. What is the area of the triangle formed by this tangent line and x- and y- axes?
1) Find the derivative of the function
y' = e2-x (-1)
y' = -e2-x
2) Find the slope tangent line at x =1
-e2-1=-e
3)Find the equation of the tangent line and solve for y
y - e = -e(x-1)
y=-ex +e +e
y=-ex + 2e
4) You know that the tangent line intersects both axes. Set the y = 0 to find the x coordinate of the x intercept
0=-ex + 2e
2e = ex
x=2
4) Set the x = 0 to find the y coordinate for the y intercept.
y=-e(0)+ 2e
y = 2e
5) The length of the y intercept to the origin is 2e (height) and the length of the x intercept to the origin is 2(base). Find the area of the triangle
A = 4e/2
A = 2e
How do you use natural log differentiation?
1) Take natural log of both sides of equation y = f(x) and use laws of logs to simplify
2)Differentiate with respect to x
3)Solve for y'
4)Replace y with f(x)
Differentiate y = (ln2x)(ln9x)
y = (ln2x)(ln9x)
lny = ln(9x)(ln(ln2x))
(1/y)(dy/dx) = ln9x(1/ln(2x))(2/2x) + (ln(ln2x))(9/9x)
(1/y)(dy/dx) = (2ln(9x)/2xln(2x)) + (9ln(ln(2x)/x)
Final Answer:
(dy/dx) = (ln2x)(ln9x) ((2ln(9x)/2xln(2x)) + (9ln(ln(2x)/x)/ xln(2x))
Explanation:
Take the natural log of each side, drop the exponent, differentiate, and simplify.
Differentiate (d/dx)(ex + e-x) / (ex + e-x)
(d/dx)(ex + e-x) / (ex + e-x)
(dy/dx) = ((ex + e-x)(ex + e-x) - (ex + e-x)(ex + e-x) / ((ex + e-x)2
(dy/dx) = (e2x + e0 + e0 + e-2x - e2x +e0 + e0 + e2x) / ((ex + e-x)2)
Final Answer:
(dy/dx) = 4/ ((ex + e-x)2)
Explanation: Use the quotient rule to differentiate. Remember that e0 = 1. Simplify by adding like terms together.
y = 3(3x+6) tanx
y' = 3(3x+6)(sec2x) + tanx(ln3)(3(3x+6)(3)
y' = 3(3x+6)(sec2x + 3tanx(ln3))
Explanation:
1)Use derivative of au rules to differentiate
2) Simplify by factoring out 3(3x+6)
Write the equation of the horizontal tangent(s) to the curve x2 + 3y - lnx +y2 = x
(d/dx)(x2 + 3y - lnx + y2) = (d/dx)(x)
2x + 3(dy/dx) - (1/x) + 2y(dy/dx) = 1
3(dy/dx) + 2y(dy/dx) = 1 + (1/x) -2x
(dy/dx)(3+2y) = (x+1-2x2)/x
(dy/dx)=(-2x2 +x +1)/(x(3+2y))
(-2x2 +x +1)/(x(3+2y)) = 0
(-2x2 +x +1) = 0
(-2x-1)(x-1) = 0
x=-(1/2) x=1
(x=-(1/2) will not work because ln(-1/2) does not exist)
12 + 3y - ln(1) + y2 = 1
3y + y2 = 0
y(3 + y) = 0
y = 0 y = -3
The horizontal tangent cannot be y = 0, so y = -3 is the horizontal tangent.
Explanation:
1)Use implicit differentiation and solve for (dy/dx). This is the derivative.
2)Set this equation equal to 0 and solve for x to find the point where the horizontal tangent is at
3) Your answer is x=-(1/2) x=1, but x=-(1/2) will not work because ln(-1/2) does not exist
4) Plug in x=1 to the original equation. You get y = -3. This is the equation of the horizontal tangent.
The log function with base a, where a>0 and a does not equal 1 is
Y = logax if and only if x = a
Differentiate g(x) = (17x)(cos2x)
y = (17x)(cos2x)
lny = ln((17x)(cos2x)
lny = (cos2x)(ln(17x))
(1/y)(dy/dx) = (cos2x)(17/17x) + ln(17x)(-sin2x)(2)
Final Answer:
(dy/dx) = 17x(cos2x)( (cos2x/x) - 2(ln17x)(sin2x) )
Explanation:
Take the natural log of both sides, drop the exponent, differentiate, and solve for (dy/dx).
Differentiate f(x) = ln(3exx4)
f'(x) = (1/3(exx4) (3(ex(4x3) + x4(ex))
f'(x) = (3ex(4x3 + x4))/(3exx4)
f'(x) = (4x3 + x4)/x4
f'(x) = x3(4+x)/(x4)
f'(x) = (4+x)/x
Final Answer:
f'(x) = (4/x) + 1
Explanation(using the chain rule):
1)Differentiate the outside function(which is an ln function so remember that (d/dx)lnx = (1/x))
2)Differentiate the inside function using the product rule
3)Factor out the common terms
4)Simplify
y = logcosx
y' = ((1/(ln10 cosx))(-sinx)
Answer:
y'= (-sinx)/(ln10 cosx)
Explanation:
1) Use the formula to differentiate
2)Simplify
The slope of the line tangent to the graph of the graph of exy=2 at the point where x =1 is
exy=2
ln(exy) = ln2
xy = ln2
y = (ln2/x)
y ' = (x(0) - ln2)/x2
y' = -(ln2)/x2
y'(1) = -(ln2)/1
y'(1) = ln2
The slope of the tangent line is -ln2
Explanation:
1)Differentiate exy=2 by taking the ln of both sides
2)lne = 1, so you are left with xy = ln2
3)Solve for y
4)Differentiate
5)Plus in x =1 to find the derivative at this point, which is the slope of the tan line
How do you apply the change of base formula when differentiating?
(d/dx)(logax) = (1/lna(x))
(d/dx)(logau) = (1/lna(u))(du/dx)
f(x) = log3(5x-4)(2x)
f(x) = log3(5x-4)(2x)
f'(x) = (1/(ln3(5x-4)(2x)) (2x(5) -(5x-4)(2))/(4x2)
f'(x) = (10x-10x + 8)/(ln3(5x-4)4x2)
f'(x) = 8/(ln3(5x-4)4x2)
Final Answer:
f'(x) = 4/(ln3(5x-4)x)
Explanation:
Use the change of base formula to differentiate and then simplify.
If f(x) = (41-5x)/(2-3x)
f'(x) = (2-3x)(41-5x)ln4(-5)-41-5x(-3)/(2-3x)2
f'(x) = -41-5x(5(2-3x)ln4)-3)/(2-3x)2
Explanation:
Use the quotient rule to differentiate. Remember the rule for differentiating a constant with an exponential function (aulna(du/dx))
Differentiate y = log2(x2/x-1)
y = 2log2x - log2(x-1)
y' = 2(1/ln2)(1/x)(1) - (1/ln2)(1/x-1)(1)
y' = (2/xln2) - (1/ln2)(1/x-1)
y' = (1/ln2)((2/x) - (1/(x-1))
Answer:
y' = (1/ln2)((x-2)/x(x-1))
Explanation:
1)Use log properties to expand
2)Bring down the exponent
3)Differentiate
4)Simplify
Find where the tangent line is horizontal to the function f(x) = (2x-4)ex. What is the equation of the tangent line.
f'(x) = (2x-4)ex + ex(2)
f'(x) = 2ex((x-2) + 1)
f'(x) = 2ex(x-1)
2ex(x-1) = 0
(x-1) = 0
x=1
When x=1, the tangent line is horizontal to the function.
y = (2(1)-4)e(1)
y = -2e
The equation of the tangent line is y = -2e
How do you differentiate exponential functions?
(d/dx)eu = eu(du/dx)
(d/dx)au = aulna(du/dx)