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Select the equivalent expression. (x4⋅y4)3=?(x^4 \cdot y^4)^3 = ?\newlineChoose 11 answer:\newline(Choice A) x24⋅y24x^{24} \cdot y^{24}\newline(Choice B) x12⋅y12x^{12} \cdot y^{12}\newline(Choice C) x7⋅y7x^7 \cdot y^7\newline(Choice D) x64⋅y64x^{64} \cdot y^{64}

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Find higher derivatives of logarithmic and power functionsright chevron icon

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h(x)=(5−6x)5h′(x)=? \begin{aligned} h(x) & =(5-6 x)^{5} \\ h^{\prime}(x) & =? \end{aligned} \newlineChoose 11 answer:\newline(A) −6x5+5x4(5−6x) -6 x^{5}+5 x^{4}(5-6 x) \newline(B) −30(5−6x)4 -30(5-6 x)^{4} \newline(C) (−6)5 (-6)^{5} \newline(D) 5(5−6x)4 5(5-6 x)^{4}
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Find d2dx2[2sin⁡(−4x−3)] \frac{d^{2}}{d x^{2}}[2 \sin (-4 x-3)] .
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f(x)=x11f′(x)= \begin{array}{l}f(x)=x^{11} \\ f^{\prime}(x)=\end{array}
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Posted 9 months ago

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f′(x)=5ex and f(7)=40+5e7.f(0)=□ \begin{array}{l}f^{\prime}(x)=5 e^{x} \text { and } f(7)=40+5 e^{7} . \\ f(0)=\square\end{array}
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Posted 9 months ago

Question
f′(x)=12x2−6x+2 and f(−1)=3.f(2)= \begin{array}{l}f^{\prime}(x)=12 x^{2}-6 x+2 \text { and } f(-1)=3 . \\ f(2)=\end{array}
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Posted 9 months ago

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f′(x)=−27ex and f(6)=36−27e6.f(0)=□ \begin{array}{l}f^{\prime}(x)=-27 e^{x} \text { and } f(6)=36-27 e^{6} . \\ f(0)=\square\end{array}
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Posted 9 months ago

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f′(x)=−5ex and f(3)=22−5e3.f(0)=□ \begin{array}{l}f^{\prime}(x)=-5 e^{x} \text { and } f(3)=22-5 e^{3} . \\ f(0)=\square\end{array}
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Posted 9 months ago

Question
T(x)=ex2+8xx4+7,T′(x)=T(x) = \frac{e^{x^2 + 8x}}{\sqrt{x^4 + 7}},\quad T'(x) =
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Posted 8 months ago

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F(x)=4x2−5x F(x)=4 x^{2}-5 x \newlinef(x)=F′(x)∫14f(x)dx= \begin{array}{l}f(x)=F^{\prime}(x) \\ \int_{1}^{4} f(x) d x=\end{array}
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Posted 7 months ago

Question
G(x)=(x+3)2g(x)=G′(x)∫−17g(x)dx= \begin{array}{l}G(x)=(x+3)^{2} \\ g(x)=G^{\prime}(x) \\ \int_{-1}^{7} g(x) d x=\end{array}
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