Differentiate Both Sides of the Equation Practice Questions

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Differentiate Both Sides of the Equation Practice Questions

Differentiate with respect to t.

y = (x³ + x – 1)⁵
y⁴ – 3x² = cos(y)
y³ – y = 3x⁴ – 10x² + 3x + 1
√x + √y = 10x³ – 7x
In(y) + e^x =x²y²
5x² + 2x + 1 = w² + 7
A² +B² = C²
A = 4π r²
C = 2π r
A =

Suppose y² + 3y = 6 – 4x³ and = 5. What is when x = –1 and y = 2?
Suppose xy² = x² + 3. What is = 8 , x = 3 , and y = –2?
Let K + e^L + L + I² . If = 5 and = 4, what is when L = 0 and 1= 3?
Suppose A³ = B² + 4C², = 8, and = –2. What is when A = 2, B = 2, and C = 1?
Suppose A = I² + 6R. If I increases by 4 feet per minute and R increases by 2 square feet every minute, how fast is A changing when 1= 20?
Suppose . Every hour, K increases by 2. How fast is R changing when K = 3 and ?
The height of a triangle increases by 2 feet every minute while its base shrinks by 6 feet every minute. How fast is the area changing when the height is 15 feet and the base is 20 feet?
The surface area of a sphere with radius r is A = 4π r² . If the radius is decreasing by 2 inches every minute, how fast is the surface area shrinking when the radius is 20 inches?
A circle increases in area by 20 square feet every hour. How fast is the radius increasing when the radius is 4 feet?
The volume of a cube grows by 1,200 square inches every minute. How fast is each side growing when each side is 10 inches?
The height of a triangle grows by 5 inches each hour. The area is increasing by 100 square inches each hour. How fast is the base of the triangle increasing when the height is 20 inches and the base is 12 inches?
One end of a 10-foot long board is lifted straight off the ground at 1 foot per second (see figure below). How fast will the other end drag along the ground after 6 seconds?
A kite is 100 feet off the ground and moving horizontally at 13 feet per second (see figure below). How quickly must the string be let out when the string is 260 feet long?

Solutions

= 5(x³ + x – 1)⁴ · (3x² + )
4y³ – 6x = –sin(y)
3y² – = 12x³ – 20x + 3
· + · = 30x² · – 7 ·
· + e^x · = 2x · y² + 2y · · x²
10x · + 2 · = 2w ·
= x · + y · – ·
2A · + 2B · = 2C ·
= 4 π r² ·
= 8 π r ·
= 2 π ·
= · · h + · b
= –
= –
= 24
= 28
Because = 172 , A is increasing at the rate of 172 square feet per minute when I = 20.
= –, so R is decreasing at the rate of per hour at this instant.
= –25, so the area is decreasing at the rate of 25 square feet per minute.
= –320 π , so the area shrinks by 320 πsquare inches per minute.
= , so the radius grows at ≈ 0.796 feet per hour.
V = s³, so = 3s² , thus = 4 when = 1,200 and s = 10. Each side is growing at the rate of 4 inches per minute.
= 7, so the base increases at the rate of 7 inches per hour.
If the height is y and the base is x, then x² + y² = 10² and 2x + 2y = 0. After 6 seconds, y = 6 and x² + 6² = 100, so x = 8. Because = 1,2(8) + 2(6)(1) = 0, so = – .The end of the board is moving at the rate of of a foot each hour along the ground.
If the base is x and the hypotenuse (length of the string) is 5, then x² + 100² = s² . Using this, when s = 260, must be 240. Because = 13, we can calculate that = 12. Thus, the string must be let out at 12 feet per second.