practice exercises\n15 - 24. for each of the following composite functions, find an inner function u = g(x)…

practice exercises\n15 - 24. for each of the following composite functions, find an inner function u = g(x) and an outer function y = f(u) such that y = f(g(x)). then calculate dy/dx.\n15. y=(3x + 7)^{10}\n16. y=(5x^{2}+11x)^{4/3}\n17. y = sin^{5}x\n18. y = sin x^{5}\n19. y = \\sqrt{x^{2}+1}\n20. y = \\sqrt{7x - 1}\n21. y = e^{4x^{2}+1}\n22. y = e^{\\sqrt{x}}\n23. y = tan 5x^{2}\n24. y = sin\\frac{x}{4}

practice exercises\n15 - 24. for each of the following composite functions, find an inner function u = g(x) and an outer function y = f(u) such that y = f(g(x)). then calculate dy/dx.\n15. y=(3x + 7)^{10}\n16. y=(5x^{2}+11x)^{4/3}\n17. y = sin^{5}x\n18. y = sin x^{5}\n19. y = \\sqrt{x^{2}+1}\n20. y = \\sqrt{7x - 1}\n21. y = e^{4x^{2}+1}\n22. y = e^{\\sqrt{x}}\n23. y = tan 5x^{2}\n24. y = sin\\frac{x}{4}

Answer

Explanation:

Step1: Identify inner and outer functions for (y=(3x + 7)^{10})

Let (u = g(x)=3x + 7) (inner - function) and (y = f(u)=u^{10}) (outer - function).

Step2: Find (\frac{du}{dx}) and (\frac{dy}{du})

(\frac{du}{dx}=\frac{d}{dx}(3x + 7)=3), (\frac{dy}{du}=\frac{d}{du}(u^{10}) = 10u^{9}).

Step3: Use the chain - rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx})

Substitute (u = 3x+7) into (\frac{dy}{du}), then (\frac{dy}{dx}=10(3x + 7)^{9}\cdot3=30(3x + 7)^{9}).

Answer:

For (y=(3x + 7)^{10}), (u = 3x + 7), (y = u^{10}), (\frac{dy}{dx}=30(3x + 7)^{9})

(We can follow the same steps for the other functions: For (y=(5x^{2}+11x)^{\frac{4}{3}}): Let (u = 5x^{2}+11x), (y = u^{\frac{4}{3}}) (\frac{du}{dx}=10x + 11), (\frac{dy}{du}=\frac{4}{3}u^{\frac{1}{3}}) (\frac{dy}{dx}=\frac{4}{3}(5x^{2}+11x)^{\frac{1}{3}}(10x + 11))

For (y=\sin^{5}x): Let (u=\sin x), (y = u^{5}) (\frac{du}{dx}=\cos x), (\frac{dy}{du}=5u^{4}) (\frac{dy}{dx}=5\sin^{4}x\cos x)

For (y = \sin x^{5}): Let (u=x^{5}), (y=\sin u) (\frac{du}{dx}=5x^{4}), (\frac{dy}{du}=\cos u) (\frac{dy}{dx}=5x^{4}\cos(x^{5}))

For (y=\sqrt{x^{2}+1}=(x^{2}+1)^{\frac{1}{2}}): Let (u=x^{2}+1), (y = u^{\frac{1}{2}}) (\frac{du}{dx}=2x), (\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}) (\frac{dy}{dx}=\frac{2x}{2\sqrt{x^{2}+1}}=\frac{x}{\sqrt{x^{2}+1}})

For (y=\sqrt{7x - 1}=(7x - 1)^{\frac{1}{2}}): Let (u = 7x-1), (y = u^{\frac{1}{2}}) (\frac{du}{dx}=7), (\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}) (\frac{dy}{dx}=\frac{7}{2\sqrt{7x - 1}})

For (y = e^{4x^{2}+1}): Let (u = 4x^{2}+1), (y = e^{u}) (\frac{du}{dx}=8x), (\frac{dy}{du}=e^{u}) (\frac{dy}{dx}=8xe^{4x^{2}+1})

For (y = e^{\sqrt{x}}): Let (u=\sqrt{x}=x^{\frac{1}{2}}), (y = e^{u}) (\frac{du}{dx}=\frac{1}{2\sqrt{x}}), (\frac{dy}{du}=e^{u}) (\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{2\sqrt{x}})

For (y=\tan(5x^{2})): Let (u = 5x^{2}), (y=\tan u) (\frac{du}{dx}=10x), (\frac{dy}{du}=\sec^{2}u) (\frac{dy}{dx}=10x\sec^{2}(5x^{2}))

For (y=\sin\frac{x}{4}): Let (u=\frac{x}{4}), (y=\sin u) (\frac{du}{dx}=\frac{1}{4}), (\frac{dy}{du}=\cos u) (\frac{dy}{dx}=\frac{1}{4}\cos\frac{x}{4}))