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2"},"content":{"rendered":""},"excerpt":{"rendered":""},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":10,"niveau_difficulte":"Intermediate","categorie":"Differentiation","nombre_questions":10,"score_minimum":70,"ordre_questions":"Fixe","questions":[{"id":"cmpz9ra0m07vujgllsf5scj2r","texte_question":"

If $$y=\\sin(2x+5)$$, find $$\\frac{dy}{dx}$$.

","type_question":"QCM","explication":"

Differentiate using the chain rule.

The derivative of $$\\sin(u)$$ is $$\\cos(u)\\cdot u'$$.

Let:

$$u=2x+5$$

Differentiate:

$$\\frac{du}{dx}=2$$

Apply the chain rule:

$$\\frac{dy}{dx}=\\cos(2x+5)\\cdot2$$

Simplify:

$$\\frac{dy}{dx}=2\\cos(2x+5)$$

Final answer: $$2\\cos(2x+5)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2\\sin(2x+5)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2\\cos(2x+5)$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-2\\sin(2x+5)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\cos(2x+5)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra0p07w0jglluchhoo73","texte_question":"

If $$y=(x+3)\\ln(x+2)$$, find $$\\frac{dy}{dx}$$.

","type_question":"QCM","explication":"

Differentiate using the product rule.

Let:

$$u=x+3$$

$$v=\\ln(x+2)$$

Differentiate each function.

$$\\frac{du}{dx}=1$$

$$\\frac{dv}{dx}=\\frac{1}{x+2}$$

Apply the product rule:

$$\\frac{dy}{dx}=u'v+uv'$$

$$\\frac{dy}{dx}=1\\cdot\\ln(x+2)+(x+3)\\cdot\\frac{1}{x+2}$$

Simplify:

$$\\frac{dy}{dx}=\\ln(x+2)+\\frac{x+3}{x+2}$$

Final answer: $$\\ln(x+2)+\\frac{x+3}{x+2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\ln(x+3)+\\frac{x+3}{x+2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{x+2}+\\frac{x+3}{x+2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\ln(x+2)+\\frac{x+3}{x+2}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\ln(x+2)+\\frac{x+2}{x+3}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra0r07w6jgllpqqp259j","texte_question":"

If $$f(x)=1+\\cos(4x)$$, find $$f'(x)$$.

","type_question":"QCM","explication":"

Differentiate term by term.

The derivative of the constant $$1$$ is $$0$$.

Differentiate $$\\cos(4x)$$ using the chain rule.

The derivative of $$\\cos(u)$$ is $$-\\sin(u)\\cdot u'$$.

Let:

$$u=4x$$

Differentiate:

$$\\frac{du}{dx}=4$$

Apply the chain rule:

$$\\frac{d}{dx}\\cos(4x)=-\\sin(4x)\\cdot4$$

Combine results:

$$f'(x)=-4\\sin(4x)$$

Final answer: $$-4\\sin(4x)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-4\\cos(4x)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\sin(4x)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4\\sin(4x)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-4\\sin(4x)$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra0u07wcjgll6f9n0rcq","texte_question":"

If $$y=\\sin^{-1}(2x)+\\sqrt{1-4x^2}$$, find $$\\frac{dy}{dx}$$.

","type_question":"QCM","explication":"$18","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{4x-2}{\\sqrt{1-4x^2}}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{2-4x}{1-4x^2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{2+4x}{\\sqrt{1-4x^2}}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{2-4x}{\\sqrt{1-4x^2}}$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra0w07wijgll812yplns","texte_question":"

If $$u=\\ln\\sqrt{x^2+4x+5}$$, find $$\\frac{du}{dx}$$.

","type_question":"QCM","explication":"

Rewrite the expression using logarithm properties:

$$u=\\ln\\left((x^2+4x+5)^{\\frac{1}{2}}\\right)$$

$$u=\\frac{1}{2}\\ln(x^2+4x+5)$$

Differentiate using the chain rule:

$$\\frac{du}{dx}=\\frac{1}{2}\\cdot\\frac{1}{x^2+4x+5}\\cdot\\frac{d}{dx}(x^2+4x+5)$$

Differentiate the inside:

$$\\frac{d}{dx}(x^2+4x+5)=2x+4$$

Substitute:

$$\\frac{du}{dx}=\\frac{1}{2}\\cdot\\frac{2x+4}{x^2+4x+5}$$

Simplify:

$$\\frac{du}{dx}=\\frac{x+2}{x^2+4x+5}$$

Final answer: $$\\frac{x+2}{x^2+4x+5}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{1}{x^2+4x+5}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{2x+4}{x^2+4x+5}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{x+2}{x^2+4x+5}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{x+2}{x^2+4x}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra0y07wojgllql5oqrkh","texte_question":"

If $$f(x)=e^x(\\sin x+\\cos x)$$, find $$f'(x)$$.

","type_question":"QCM","explication":"

Differentiate using the product rule.

Let:

$$u=e^x$$

$$v=\\sin x+\\cos x$$

Differentiate each function:

$$u'=e^x$$

$$v'=\\cos x-\\sin x$$

Apply the product rule:

$$f'(x)=u'v+uv'$$

$$f'(x)=e^x(\\sin x+\\cos x)+e^x(\\cos x-\\sin x)$$

Factor out $$e^x$$:

$$f'(x)=e^x[(\\sin x+\\cos x)+(\\cos x-\\sin x)]$$

Simplify inside the brackets:

$$\\sin x-\\sin x=0$$

$$\\cos x+\\cos x=2\\cos x$$

So:

$$f'(x)=2e^x\\cos x$$

Final answer: $$2e^x\\cos x$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2e^x\\cos x$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2e^x(\\sin x+\\cos x)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$e^x(\\sin x+\\cos x)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2e^x\\sin x$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra1107wujgllmtrn191f","texte_question":"

Let $$y=3u^3-2u^2+4u-1$$ and $$u=x^2+2x$$. Find $$\\frac{dy}{dx}$$.

","type_question":"QCM","explication":"

Use the chain rule:

$$\\frac{dy}{dx}=\\frac{dy}{du}\\cdot\\frac{du}{dx}$$

Differentiate $$y$$ with respect to $$u$$:

$$\\frac{dy}{du}=9u^2-4u+4$$

Differentiate $$u$$ with respect to $$x$$:

$$\\frac{du}{dx}=2x+2$$

Apply the chain rule:

$$\\frac{dy}{dx}=(9u^2-4u+4)(2x+2)$$

Substitute $$u=x^2+2x$$:

$$\\frac{dy}{dx}=(9(x^2+2x)^2-4(x^2+2x)+4)(2x+2)$$

Simplify $$(x^2+2x)^2$$:

$$(x^2+2x)^2=x^4+4x^3+4x^2$$

Substitute:

$$9(x^4+4x^3+4x^2)-4x^2-8x+4$$

$$=9x^4+36x^3+36x^2-4x^2-8x+4$$

$$=9x^4+36x^3+32x^2-8x+4$$

Multiply by $$(2x+2)$$:

$$\\frac{dy}{dx}=(9x^4+36x^3+32x^2-8x+4)(2x+2)$$

Final answer:

$$\\frac{dy}{dx}=(9x^4+36x^3+32x^2-8x+4)(2x+2)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(9x^4+36x^3+28x^2-8x+4)(2x+2)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(9x^4+36x^3+32x^2-8x+4)(2x+2)$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$18x^4+72x^3+64x^2-16x+8$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2x(9x^4+36x^3+32x^2-8x+4)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra1307x0jgllou3q5eyj","texte_question":"

If $$y=\\cos(ax+b)$$, with $$a$$ and $$b$$ constants, find $$\\frac{dy}{dx}$$.

","type_question":"QCM","explication":"

Differentiate using the chain rule.

The derivative of $$\\cos(u)$$ is $$-\\sin(u)\\cdot u'$$.

Let:

$$u=ax+b$$

Differentiate:

$$\\frac{du}{dx}=a$$

Apply the chain rule:

$$\\frac{dy}{dx}=-\\sin(ax+b)\\cdot a$$

Simplify:

$$\\frac{dy}{dx}=-a\\sin(ax+b)$$

Final answer: $$-a\\sin(ax+b)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\sin(ax+b)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-a\\cos(ax+b)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$a\\sin(ax+b)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-a\\sin(ax+b)$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra1607x6jglliul1o301","texte_question":"

If $$g(x)=b\\sin(kx)$$ with $$b$$ and $$k$$ constants, find $$g'(x)$$ and $$g''(x)$$.

","type_question":"QCM","explication":"

Differentiate $$g(x)=b\\sin(kx)$$.

The derivative of $$\\sin(u)$$ is $$\\cos(u)\\cdot u'$$.

Let:

$$u=kx$$

Then:

$$\\frac{du}{dx}=k$$

So:

$$g'(x)=b\\cos(kx)\\cdot k$$

Simplify:

$$g'(x)=bk\\cos(kx)$$

Differentiate again to find $$g''(x)$$.

The derivative of $$\\cos(u)$$ is $$-\\sin(u)\\cdot u'$$.

Using $$u=kx$$ and $$u'=k$$:

$$g''(x)=bk(-\\sin(kx))\\cdot k$$

Simplify:

$$g''(x)=-bk^2\\sin(kx)$$

Final answer: $$g'(x)=bk\\cos(kx)$$, $$g''(x)=-bk^2\\sin(kx)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$g'(x)=bk\\cos(kx)$$, $$g''(x)=-bk^2\\sin(kx)$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g'(x)=bk^2\\cos(kx),\\ g''(x)=-b\\sin(kx)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g'(x)=bk\\sin(kx),\\ g''(x)=bk^2\\cos(kx)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g'(x)=b\\cos(kx),\\ g''(x)=-bk\\sin(kx)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9ra1807xcjgll6dn50x2k","texte_question":"

If $$y=\\ln(ax+b)$$, where $$a$$ and $$b$$ are constants, find $$\\frac{dy}{dx}$$.

","type_question":"QCM","explication":"

Differentiate using the chain rule.

The derivative of $$\\ln(u)$$ is $$\\frac{u'}{u}$$.

Let:

$$u=ax+b$$

Differentiate:

$$\\frac{du}{dx}=a$$

Apply the chain rule:

$$\\frac{dy}{dx}=\\frac{a}{ax+b}$$

Final answer: $$\\frac{a}{ax+b}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{b}{ax+b}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{a}{x+b}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{a}{ax+b}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{ax+b}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["differentiation"],"date":"2026-06-04T09:04:33.907Z","modified":"2026-06-04T09:04:33.907Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L19",null,{"targetType":"quiz","targetSlug":"differentiation-quiz-2"}]}]]}]}]]

Quiz 2

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