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I found an answer here: https://brainly.in/question/3822188

Let, $ y = \sqrt {\text{sin}\sqrt {x}} $

or, y = (sin√x)^(1/2)

Now, differentiating with respect to x, we get

dy/dx

= 1/2 (sin√x)^(1/2 - 1) d/dx (sin√x)

= 1/2 (sin√x)^(-1/2) (cos√x) d/dx (√x)

= 1/2 1/√(sin√x) (cos√x) d/dx {x^(1/2)}

= 1/2 1/√(sin√x) (cos√x) 1/2 x^(1/2 - 1)

= 1/2 1/√(sin√x) (cos√x) 1/2 x^(-1/2)

= (1/2 × 1/2) √(sin√x) (cos√x) 1/(√x)

= 1/4 √(sin√x) (cos√x) 1/(√x)

What I don't understand is how 1/2 1/√(sin√x) (cos√x) 1/2 x^(-1/2) turned to (1/2 × 1/2) √(sin√x) (cos√x) 1/(√x)? Which formula is applied here? Also, it would be helpful if someone wrote it in LaTex which will help understand better.

MANI
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hmmmm
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  • 1/√(sin√x) turns to √(sin√x)? Is it because 1/√(sin√x) = √(sin√x)? – hmmmm Dec 04 '19 at 12:20
  • So you're wondering why $$ \frac{1}{2} \frac{1}{\sqrt{\sin{\sqrt{x}}}} \cdot \cos{\sqrt{x}} \cdot \frac{1}{2} x^{-1/2} $$ is equal to $$ \frac{1}{2} \cdot \frac{1}{2} \sqrt{\sin{\sqrt{x}}} \cdot \cos{\sqrt{x}} \cdot \frac{1}{\sqrt{x}} $$ or something like this? There are three things that changed: one instance of $\frac{1}{2}$ was moved to the beginning and then it was noted that $$ x^{-1/2} = \frac{1}{\sqrt{x}} $$ Also, remember that for any $a \in \mathbb{R}$, $$ \frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{\sqrt{a} \sqrt{a}} = \frac{\sqrt{a}}{a} $$ – Matti P. Dec 04 '19 at 12:21
  • @MattiP. I got that but I don't understand the next line. The 2nd line from last to first. – hmmmm Dec 04 '19 at 12:23
  • @MattiP. $$\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{\sqrt{a} \sqrt{a}} = \frac{\sqrt{a}}{a}$$

    Wouldn't that result in [√(sin√x)]/sin√x instead of just √(sin√x) ?

     – hmmmm Dec 04 '19 at 12:30
  • yes it would. I'm a bit blinded currently by the long expressions, I don't see where the error is ... – Matti P. Dec 04 '19 at 12:32
  • @hmmmm you are right in your last comment. There is indeed a term missing, the one you mention – Mihail Dec 04 '19 at 13:27
  • I don't get why people here answer to the problem in the title instead of reading the whole post and addressing the actual question of OP. We don't solve problems just to get to the answer but rather to learn how to solve stuff and how not to solve. I apologize for rudeness. – Mihail Dec 04 '19 at 13:33

3 Answers3

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The mistake in the solution that you found is

= (1/2 × 1/2) $\color{red}{√(\sin√x)}$ (cos√x) 1/(√x)

It should be

= (1/2 × 1/2) $\color{blue}{1/{√(\sin√x)}}$ (cos√x) 1/(√x)

When solutions are typed out from left to right like the one that you found, it is very easy to misplace the numerators and denominators if you are writing a product of multiple fractions.

Andrew Chin
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    One thing that is ridiculously helpful is that, once you have learned that $$\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{1/2}=1/2\ x^{-1/2}=1/(2\sqrt{x}),$$ it is much less confusing just to remember that $$\frac{d}{dx}\sqrt{x}=\frac1{2\sqrt{x}}.$$ – Andrew Chin Dec 04 '19 at 14:18
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Writing $$f(x)=(\sin(x^{1/2})^{1/2}$$ we get by the chain rule $$\frac{1}{2}(\sin(x^{1/2})^{-1/2}\times \cos(x^{1/2})\times 1/2x^{-1/2}$$

Dr. Sonnhard Graubner
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  • I am not sure if the answer I found is correct or not. I don't get the 2nd line from last to first. – hmmmm Dec 04 '19 at 12:28
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We know that

$$f(x)=\sqrt x \implies f'(x)=\frac1{2\sqrt x}$$

therefore by chain rule

$$f(g(x))=\sqrt {g(x)} \implies f'(x)=\frac{g'(x)}{2\sqrt {g(x)}}$$

with

$$g(x)=\sin \sqrt x \implies g'(x)=\cos \sqrt x \cdot \frac1{2\sqrt x}$$

user
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