Author Topic: hyunah's Specialist Questions (Read 3403 times) Tweet Share

hyunah · « **Reply #15 on:** August 24, 2014, 05:13:04 pm »

how do i differentiate:
tan^-1(2sqrt(x/15)

the ans says: sqrt(15)/[sqrt(x)(4x+15)

polar1 · « **Reply #16 on:** August 24, 2014, 05:44:21 pm »

Quote from: hyunah on August 24, 2014, 05:13:04 pm

how do i differentiate:
tan^-1(2sqrt(x/15)

the ans says: sqrt(15)/[sqrt(x)(4x+15)

Let $y=\arctan\left(2\sqrt\frac{x}{15} \right) = \arctan\left(\frac{2}{\sqrt{15}}\sqrt{x} \right)$ . Then, let $u=\frac{2\sqrt{x}}{\sqrt{15}}$ so that $y=\arctan(u)$ and $\frac{\mathrm{d}y}{\mathrm{d}u} = \frac{1}{1+u^2}$ while $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{\sqrt{15}}\times\frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{15}\sqrt{x}}$ . Then, using the chain rule, we have $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1+u^2}\times\frac{1}{\sqrt{15}\sqrt{x}} = \frac{1}{1+(4x/15)}\times\frac{1}{\sqrt{15}\sqrt{x}}$ .

Multiply the numerator and denominator of the first fraction by 15 to get $\frac{15}{15 + 4x}\times\frac{1}{\sqrt{15}\sqrt{x}} = \frac{\sqrt{15}}{\sqrt{x}(15+4x)}$

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Author Topic: hyunah's Specialist Questions (Read 3403 times) Tweet Share

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