1. differentiate the following: 3 marks each\n(a) ( f(t)=cos left(sin ^{-1} 4 t\right) )\n(b) ( g(t)=ln…

Explanation:

Step1: Differentiate (f(t)=\cos(\sin^{-1}4t))

Let (u = \sin^{-1}4t). Then (f(t)=\cos u). By the chain - rule (\frac{df}{dt}=\frac{df}{du}\cdot\frac{du}{dt}). We know that (\frac{d}{du}(\cos u)=-\sin u) and (\frac{d}{dt}(\sin^{-1}4t)=\frac{4}{\sqrt{1-(4t)^{2}}}). Since (\sin(\sin^{-1}x)=x) for (x\in[- 1,1]), (f(t)=\cos(\sin^{-1}4t)=\sqrt{1 - 16t^{2}}) (using the identity (\cos(\sin^{-1}x)=\sqrt{1 - x^{2}}) for (x\in[-1,1])). Differentiating (y = \sqrt{1-16t^{2}}=(1 - 16t^{2})^{\frac{1}{2}}) using the chain - rule: (\frac{dy}{dt}=\frac{1}{2}(1 - 16t^{2})^{-\frac{1}{2}}\cdot(-32t)=-\frac{16t}{\sqrt{1 - 16t^{2}}})

Step2: Differentiate (g(t)=\ln(\sqrt{t^{2}+4}))

First, simplify (g(t)) using the property (\ln(a^{b})=b\ln a). So (g(t)=\frac{1}{2}\ln(t^{2}+4)). By the chain - rule, if (y=\frac{1}{2}\ln u) where (u = t^{2}+4), (\frac{dy}{dt}=\frac{1}{2}\cdot\frac{1}{u}\cdot\frac{du}{dt}). Since (\frac{du}{dt}=2t), then (\frac{dg}{dt}=\frac{t}{t^{2}+4})

Step3: Differentiate (h(t)=3^{t}\log_{2}(t^{2}+1))

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 3^{t}) and (v=\log_{2}(t^{2}+1)). We know that (\frac{d}{dt}(3^{t})=3^{t}\ln3) and (\frac{d}{dt}(\log_{2}(t^{2}+1))=\frac{2t}{(t^{2}+1)\ln2}) (using the formula (\frac{d}{dt}(\log_{a}u)=\frac{1}{u\ln a}\cdot\frac{du}{dt})). So (h^\prime(t)=3^{t}\ln3\log_{2}(t^{2}+1)+3^{t}\cdot\frac{2t}{(t^{2}+1)\ln2}=3^{t}\left(\ln3\log_{2}(t^{2}+1)+\frac{2t}{(t^{2}+1)\ln2}\right))

Step4: Differentiate (y(t)=e^{-2t}\tan^{-1}t)

Using the product rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = e^{-2t}) and (v=\tan^{-1}t). (\frac{d}{dt}(e^{-2t})=-2e^{-2t}) and (\frac{d}{dt}(\tan^{-1}t)=\frac{1}{1 + t^{2}}). So (y^\prime(t)=-2e^{-2t}\tan^{-1}t+\frac{e^{-2t}}{1 + t^{2}}=e^{-2t}\left(\frac{1}{1 + t^{2}}-2\tan^{-1}t\right))

Step5: Differentiate (x(t)=\cos(te^{t}))

Let (u = te^{t}). Then (x(t)=\cos u). First, find (\frac{du}{dt}) using the product rule: (\frac{du}{dt}=e^{t}+te^{t}=(t + 1)e^{t}). By the chain - rule (\frac{dx}{dt}=-\sin u\cdot\frac{du}{dt}). Substituting (u = te^{t}) and (\frac{du}{dt}=(t + 1)e^{t}), we get (\frac{dx}{dt}=-(t + 1)e^{t}\sin(te^{t}))

Answer:

(a) (-\frac{16t}{\sqrt{1 - 16t^{2}}}) (b) (\frac{t}{t^{2}+4}) (c) (3^{t}\left(\ln3\log_{2}(t^{2}+1)+\frac{2t}{(t^{2}+1)\ln2}\right)) (d) (e^{-2t}\left(\frac{1}{1 + t^{2}}-2\tan^{-1}t\right)) (e) (-(t + 1)e^{t}\sin(te^{t}))