Porisma

Worked solution

Practise power, chain and product rules together

Differentiate (a) 2/x+6/x32/\sqrt{x}+6/\sqrt[3]{x} and (b) (x2+3)4(2x3−5)3(x^2+3)^4(2x^3-5)^3 with respect to xx.

CJC · Lecture · Self-Practice 1

(a) Convert the radicals to powers

For x>0x>0, the expression is 2x−1/2+6x−1/32x^{-1/2}+6x^{-1/3}. Differentiate to get −x−3/2−2x−4/3-x^{-3/2}-2x^{-4/3}, matching the printed answer. The square-root denominator restricts the combined real domain to x>0x>0.

How did this part go?

(b) Apply the product and chain rules

The derivative is 8x(x2+3)3(2x3−5)3+18x2(x2+3)4(2x3−5)28x(x^2+3)^3(2x^3-5)^3+18x^2(x^2+3)^4(2x^3-5)^2.

How did this part go?

(b) Factor without cancelling possible zeros

Take out 2x(x2+3)3(2x3−5)22x(x^2+3)^3(2x^3-5)^2. The remaining bracket is 4(2x3−5)+9x(x2+3)=17x3+27x−204(2x^3-5)+9x(x^2+3)=17x^3+27x-20. Thus the result is 2x(x2+3)3(2x3−5)2(17x3+27x−20)2x(x^2+3)^3(2x^3-5)^2(17x^3+27x-20), exactly the printed answer. This polynomial derivative is valid for all real xx; factoring does not exclude zeros of the factors.

How did this part go?

Common mistakes

The square and cube roots in part (a) are different; both chain factors are needed in part (b).

Solution notes

Both originals print the final answers but not intermediate working. The intermediate derivation and domain commentary here are independently authored and checked against both printed answers.

Read the hint