Practice problem
Find the derivative. λ(t)=ln(sin−1tcos−1t)\lambda(t)=\ln\bigg(\frac{\sin^{-1}t}{\cos^{-1}t}\bigg)λ(t)=ln(cos−1tsin−1t)
λ′(t)=sint+cost1−t2\lambda'(t)=\frac{\sin t+\cos t}{\sqrt{1-t^2}}λ′(t)=1−t2sint+cost
λ′(t)=cos−1t−sin−1tsin−1tcos−1t\lambda'(t)=\frac{\cos^{-1}t-\sin^{-1}t}{\sin^{-1}t\cos^{-1}t}λ′(t)=sin−1tcos−1tcos−1t−sin−1t
λ′(t)=1tan−1t(1+t2)\lambda'(t)=\frac{1}{\tan^{-1}t(1+t^2)}λ′(t)=tan−1t(1+t2)1
λ′(t)=sin−1t+cos−1t1−t2sin−1tcos−1t\lambda'(t)=\frac{\sin^{-1}t+\cos^{-1}t}{\sqrt{1-t^2}\sin^{-1}t\cos^{-1}t}λ′(t)=1−t2sin−1tcos−1tsin−1t+cos−1t
Type a concept, like “slope” or “photosynthesis”.
