Find the derivatives for each function: 1. y = cos4(sin2(x3)). 2. f(x) = sqrt(x2 + sqrt(x2 + 1)).
|Mathematics
Find the derivatives for each function: 1. y = cos4(sin2(x3)). 2. f(x) = sqrt(x2 + sqrt(x2 + 1)).
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Answer
−24x2cos3(sin2(x3))sin(sin2(x3))sin(x3)cos(x3)
Hey ThegirlGodishelping🥰💙🫠 — ready when you are.
Here are the derivatives for each function:
1. Find dxdy if y=cos4(sin2(x3)).
Step 1: Rewrite the function to clearly show the layers of the chain rule.
y=(cos(sin2(x3)))4
Step 2: Apply the chain rule repeatedly.
The outermost function is (⋅)4.
dxdy=4(cos(sin2(x3)))3⋅dxd(cos(sin2(x3)))
Step 3: Differentiate the next layer, cos(⋅).
dxdy=4cos3(sin2(x3))⋅(−sin(sin2(x3)))⋅dxd(sin2(x3))
Step 4: Differentiate the next layer, (⋅)2.
dxdy=−4cos3(sin2(x3))sin(sin2(x3))⋅2sin(x3)⋅dxd(sin(x3))
Step 5: Differentiate the next layer, sin(⋅).
dxdy=−8cos3(sin2(x3))sin(sin2(x3))sin(x3)⋅cos(x3)⋅dxd(x3)
Step 6: Differentiate the innermost layer, x3.
dxdy=−8cos3(sin2(x3))sin(sin2(x3))sin(x3)cos(x3)⋅(3x2)
Step 7: Simplify the expression.
dxdy=−24x2cos3(sin2(x3))sin(sin2(x3))sin(x3)cos(x3)
2. Find f′(x) if f(x)=x2+x2+1.
Step 1: Rewrite the function using exponents.
f(x)=(x2+(x2+1)1/2)1/2
Step 2: Apply the chain rule. The outermost function is (⋅)1/2.
f′(x)=21(x2+(x2+1)1/2)−1/2⋅dxd(x2+(x2+1)1/2)
Step 3: Differentiate the term inside the brackets.
dxd(x2+(x2+1)1/2)=2x+21(x2+1)−1/2⋅dxd(x2+1)=2x+21(x2+1)−1/2⋅(2x)=2x+x(x2+1)−1/2=2x+x2+1x=x2+12xx2+1+x=x2+1x(2x2+1+1)
Step 4: Substitute this back into the expression for f′(x).
f′(x)=2x2+x2+11⋅x2+1x(2x2+1+1)f'(x) = \frac{x(2\sqrt{x^2 + 1 + 1)}{2\sqrt{x^2 + 1}\sqrt{x^2 + \sqrt{x^2 + 1}}}}
3. Differentiate y=(x2+1)sinx with respect to x.
Step 1: Use logarithmic differentiation. Take the natural logarithm of both sides.
lny=ln((x2+1)sinx)lny=sinx⋅ln(x2+1)
Step 2: Differentiate both sides with respect to x. Use the product rule on the right side.
y1dxdy=dxd(sinx)⋅ln(x2+1)+sinx⋅dxd(ln(x2+1))y1dxdy=(cosx)ln(x2+1)+sinx⋅(x2+11⋅2x)y1dxdy=cosxln(x2+1)+x2+12xsinx
Step 3: Solve for dxdy.
dxdy=y(cosxln(x2+1)+x2+12xsinx)
Substitute y=(x2+1)sinx back into the equation.
dxdy=(x2+1)sinx(cosxln(x2+1)+x2+12xsinx)
4. Find dxdy for the equation sin(x+y)=y2cosx.
Step 1: Differentiate both sides of the equation with respect to x. Remember to use the chain rule for terms involving y and the product rule for y2cosx.
dxd(sin(x+y))=dxd(y2cosx)cos(x+y)⋅dxd(x+y)=dxd(y2)cosx+y2dxd(cosx)cos(x+y)(1+dxdy)=2ydxdycosx+y2(−sinx)cos(x+y)+cos(x+y)dxdy=2ycosxdxdy−y2sinx
Step 2: Rearrange the equation to group terms with dxdy on one side.
cos(x+y)dxdy−2ycosxdxdy=−y2sinx−cos(x+y)
Step 3: Factor out dxdy.
dxdy(cos(x+y)−2ycosx)=−y2sinx−cos(x+y)
Step 4: Solve for dxdy.
dxdy=cos(x+y)−2ycosx−y2sinx−cos(x+y)\frac{dy}{dx} = \frac{y^2 \sin x + \cos(x + y){2y \cos x - \cos(x + y)}}
5. Find the derivative of y=(2x+3)4x53x−1 using logarithmic differentiation.
Step 1: Rewrite the function using exponents.
y=(2x+3)4x5(x−1)1/3
Step 2: Take the natural logarithm of both sides.
lny=ln((2x+3)4x5(x−1)1/3)
Step 3: Use logarithm properties to expand the expression.
lny=ln(x5)+ln((x−1)1/3)−ln((2x+3)4)lny=5lnx+31ln(x−1)−4ln(2x+3)
Step 4: Differentiate both sides with respect to x.
y1dxdy=5⋅x1+31⋅x−11⋅dxd(x−1)−4⋅2x+31⋅dxd(2x+3)y1dxdy=x5+3(x−1)1⋅(1)−2x+34⋅(2)y1dxdy=x5+3(x−1)1−2x+38
Step 5: Solve for dxdy.
dxdy=y(x5+3(x−1)1−2x+38)
Substitute y=(2x+3)4x53x−1 back into the equation.
\frac{dy}{dx} = \frac{x^5 \sqrt[3]{x - 1}{(2x + 3)^4} \left(\frac{5}{x} + \frac{1}{3(x - 1)} - \frac{8}{2x + 3}\right)}
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This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Hey ThegirlGodishelping🥰💙🫠 — ready when you are. Here are the derivatives for each function: 1. Find (dy)/(dx) if y = ^4(^2(x^3)). Step 1: Rewrite the function to clearly show the layers of the chain rule. y = ((^2(x^3)))^4 Step 2: Apply the chain rule repeatedly. The outermost function is (·)^4. (dy)/(dx) = 4((^2(x^3)))^3 · (d)/(dx)((^2(x^3))) Step 3: Differentiate the next layer, (·). (dy)/(dx) = 4^3(^2(x^3)) · (-(^2(x^3))) · (d)/(dx)(^2(x^3)) Step 4: Differentiate the next layer, (·)^2. (dy)/(dx) = -4^3(^2(x^3))(^2(x^3)) · 2(x^3) · (d)/(dx)((x^3)) Step 5: Differentiate the next layer, (·). (dy)/(dx) = -8^3(^2(x^3))(^2(x^3))(x^3) · (x^3) · (d)/(dx)(x^3) Step 6: Differentiate the innermost layer, x^3. (dy)/(dx) = -8^3(^2(x^3))(^2(x^3))(x^3)(x^3) · (3x^2) Step 7: Simplify the expression. (dy)/(dx) = -24x^2 ^3(^2(x^3))(^2(x^3))(x^3)(x^3) --- 2. Find f'(x) if f(x) = sqrt(x^2 + x^2 + 1). Step 1: Rewrite the function using exponents. f(x) = (x^2 + (x^2 + 1)^1/2)^1/2 Step 2: Apply the chain rule. The outermost function is (·)^1/2. f'(x) = (1)/(2)(x^2 + (x^2 + 1)^1/2)^-1/2 · (d)/(dx)(x^2 + (x^2 + 1)^1/2) Step 3: Differentiate the term inside the brackets. (d)/(dx)(x^2 + (x^2 + 1)^1/2) = 2x + (1)/(2)(x^2 + 1)^-1/2 · (d)/(dx)(x^2 + 1) = 2x + (1)/(2)(x^2 + 1)^-1/2 · (2x) = 2x + x(x^2 + 1)^-1/2 = 2x + (x)/(sqrt(x^2 + 1)) = 2xsqrt(x^2 + 1) + xsqrt(x^2 + 1) = x(2sqrt(x^2 + 1) + 1)sqrt(x^2 + 1) Step 4: Substitute this back into the expression for f'(x). f'(x) = (1)/(2sqrt(x^2 + x^2 + 1)) · x(2sqrt(x^2 + 1) + 1)sqrt(x^2 + 1) f'(x) = x(2sqrt(x^2 + 1) + 1)2sqrt(x^2 + 1)sqrt(x^2 + x^2 + 1) --- 3. Differentiate y = (x^2 + 1)^ x with respect to x. Step 1: Use logarithmic differentiation. Take the natural logarithm of both sides. y = ((x^2 + 1)^ x) y = x · (x^2 + 1) Step 2: Differentiate both sides with respect to x. Use the product rule on the right side. (1)/(y) (dy)/(dx) = (d)/(dx)( x) · (x^2 + 1) + x · (d)/(dx)((x^2 + 1)) (1)/(y) (dy)/(dx) = ( x) (x^2 + 1) + x · ((1)/(x^2 + 1) · 2x) (1)/(y) (dy)/(dx) = x (x^2 + 1) + (2x x)/(x^2 + 1) Step 3: Solve for (dy)/(dx). (dy)/(dx) = y ( x (x^2 + 1) + (2x x)/(x^2 + 1)) Substitute y = (x^2 + 1)^ x back into the equation. (dy)/(dx) = (x^2 + 1)^ x ( x (x^2 + 1) + (2x x)/(x^2 + 1)) --- 4. Find (dy)/(dx) for the equation (x + y) = y^2 x. Step 1: Differentiate both sides of the equation with respect to x. Remember to use the chain rule for terms involving y and the product rule for y^2 x. (d)/(dx)((x + y)) = (d)/(dx)(y^2 x) (x + y) · (d)/(dx)(x + y) = (d)/(dx)(y^2) x + y^2 (d)/(dx)( x) (x + y) (1 + (dy)/(dx)) = 2y (dy)/(dx) x + y^2 (- x) (x + y) + (x + y) (dy)/(dx) = 2y x (dy)/(dx) - y^2 x Step 2: Rearrange the equation to group terms with (dy)/(dx) on one side. (x + y) (dy)/(dx) - 2y x (dy)/(dx) = -y^2 x - (x + y) Step 3: Factor out (dy)/(dx). (dy)/(dx)((x + y) - 2y x) = -y^2 x - (x + y) Step 4: Solve for (dy)/(dx). (dy)/(dx) = (-y^2 x - (x + y))/((x + y) - 2y x) (dy)/(dx) = (y^2 x + (x + y))/(2y x - (x + y)) --- 5. Find the derivative of y = x^5 [3]x - 1(2x + 3)^4 using logarithmic differentiation. Step 1: Rewrite the function using exponents. y = x^5 (x - 1)^1/3(2x + 3)^4 Step 2: Take the natural logarithm of both sides. y = (x^5 (x - 1)^1/3(2x + 3)^4) Step 3: Use logarithm properties to expand the expression. y = (x^5) + ((x - 1)^1/3) - ((2x + 3)^4) y = 5 x + (1)/(3) (x - 1) - 4 (2x + 3) Step 4: Differentiate both sides with respect to x. (1)/(y) (dy)/(dx) = 5 · (1)/(x) + (1)/(3) · (1)/(x - 1) · (d)/(dx)(x - 1) - 4 · (1)/(2x + 3) · (d)/(dx)(2x + 3) (1)/(y) (dy)/(dx) = (5)/(x) + (1)/(3(x - 1)) · (1) - (4)/(2x + 3) · (2) (1)/(y) (dy)/(dx) = (5)/(x) + (1)/(3(x - 1)) - (8)/(2x + 3) Step 5: Solve for (dy)/(dx). (dy)/(dx) = y ((5)/(x) + (1)/(3(x - 1)) - (8)/(2x + 3)) Substitute y = x^5 [3]x - 1(2x + 3)^4 back into the equation. (dy)/(dx) = x^5 [3]x - 1(2x + 3)^4 ((5)/(x) + (1)/(3(x - 1)) - (8)/(2x + 3)) What's next?