Differentiate the function y equals left parenthesis 9 x Superscript 4 Baseline minus x plus 8 right parenthesis left parenthesis negative x Superscript 5 Baseline plus 2 right parenthesis
The correct answer and explanation is:
We are asked to differentiate the function: y=(9×4−x+8)(−x5+2)y = (9x^4 – x + 8)(-x^5 + 2)
This is a product of two functions, so we will apply the Product Rule for differentiation.
Step 1: Recall the Product Rule
If: y=u(x)⋅v(x),y = u(x) \cdot v(x),
then: dydx=u′(x)v(x)+u(x)v′(x)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)
Step 2: Define the two parts
Let:
- u(x)=9×4−x+8u(x) = 9x^4 – x + 8
- v(x)=−x5+2v(x) = -x^5 + 2
Differentiate both:
- u′(x)=ddx(9×4−x+8)=36×3−1u'(x) = \frac{d}{dx}(9x^4 – x + 8) = 36x^3 – 1
- v′(x)=ddx(−x5+2)=−5x4v'(x) = \frac{d}{dx}(-x^5 + 2) = -5x^4
Step 3: Apply the Product Rule
dydx=u′(x)v(x)+u(x)v′(x)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)
Substitute: dydx=(36×3−1)(−x5+2)+(9×4−x+8)(−5×4)\frac{dy}{dx} = (36x^3 – 1)(-x^5 + 2) + (9x^4 – x + 8)(-5x^4)
Step 4: Expand both terms
First term: (36×3−1)(−x5+2)=36×3(−x5)+36×3(2)−1(−x5)−1(2)=−36×8+72×3+x5−2(36x^3 – 1)(-x^5 + 2) = 36x^3(-x^5) + 36x^3(2) -1(-x^5) – 1(2) \\ = -36x^8 + 72x^3 + x^5 – 2
Second term: (9×4−x+8)(−5×4)=9×4(−5×4)+(−x)(−5×4)+8(−5×4)=−45×8+5×5−40×4(9x^4 – x + 8)(-5x^4) = 9x^4(-5x^4) + (-x)(-5x^4) + 8(-5x^4) \\ = -45x^8 + 5x^5 – 40x^4
Step 5: Combine like terms
dydx=(−36×8+72×3+x5−2)+(−45×8+5×5−40×4)\frac{dy}{dx} = (-36x^8 + 72x^3 + x^5 – 2) + (-45x^8 + 5x^5 – 40x^4)
Group and combine:
- −36×8−45×8=−81×8-36x^8 – 45x^8 = -81x^8
- x5+5×5=6x5x^5 + 5x^5 = 6x^5
- −40×4-40x^4
- +72×3+72x^3
- −2-2
✅ Final Answer:
dydx=−81×8+6×5−40×4+72×3−2\boxed{\frac{dy}{dx} = -81x^8 + 6x^5 – 40x^4 + 72x^3 – 2}
🔍 Explanation (300 words):
To differentiate a function involving a product of two expressions, the Product Rule is the key tool. The Product Rule allows us to take the derivative of two functions multiplied together without first expanding them. It states that the derivative of y=u(x)v(x)y = u(x)v(x) is u′(x)v(x)+u(x)v′(x)u'(x)v(x) + u(x)v'(x).
Here, we define u(x)=9×4−x+8u(x) = 9x^4 – x + 8 and v(x)=−x5+2v(x) = -x^5 + 2. Each part is a polynomial, so we differentiate using basic rules:
- The derivative of 9x49x^4 is 36x336x^3,
- The derivative of −x-x is −1-1,
- Constants like 8 and 2 differentiate to zero.
Thus, u′(x)=36×3−1u'(x) = 36x^3 – 1 and v′(x)=−5x4v'(x) = -5x^4.
Next, we plug into the Product Rule formula: dydx=u′(x)v(x)+u(x)v′(x)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)
We expand both products carefully using the distributive property. After distributing and multiplying terms, we group like terms—specifically powers of xx: x8,x5,x4,x3x^8, x^5, x^4, x^3, and constants. Combining all, we simplify the result.
The final answer shows the derivative as a single polynomial, expressed in standard form with descending powers of xx. This systematic approach helps avoid errors and ensures that each component contributes correctly to the derivative.