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If $y = x^3 \sin^2 (5x)$ find $\frac{dy}{dx}$.

This question was on chain rule chapter. So, I think my answer isn't correct.

$$\frac{dy}{dx}=3x^2\sin^2 (5x) + x^3 \cos^2 (5x)$$

I got above answer. I was thinking that it was wrong. So, I had try again.

$$\frac{dy}{dx}=3x^2\sin^2 (5x) +x^3 . 2\cos (5x) . 5 = 3x^2\sin^2 (5x) + 10 x^3 \cos (5x)$$

Both of them are looking wrong to me. So, could you please say what am I missing here?


By getting the comment's help, I had tried to solve it again.

$$\frac{dy}{dx}=3x^2 \sin^2 (5x)+x^3 . 2 (\sin (5x)) . \cos (5x) . 5$$ $$=3x^2 \sin^2 (5x) + 10 x^3 \sin (5x) . \cos (5x)$$

But, the answer didn't match. What am I doing wrong again?

I had seen something right there while differentiating.

$$\frac{dy}{dx}=3x^2 \sin^2 (5x)+x^3 . 2 (\sin (5x))^{2-1} . \cos (5x)^{1-1} . 5$$

So, simply the $\cos (5x)$ should be gone. I am thinking wrong now. Or, I was thinking wrong earlier?

How Gaurang got $\frac{dy}{dx} = 3x^2\sin^2(5x)+5x^3 10\sin(10x)$? I am thinking of $5x^3 10\sin(10x)$ I think I am missing something in Trigonometric function.


So, correct answer is $$3x^2 \sin^2 (5x) + 10 x^3 \sin (5x) . \cos (5x)$$

according to this comment.

I had asked another question; Why this isn't happening?

$$\frac{dy}{dx}=3x^2 \sin^2 (5x)+x^3 . 2 (\sin (5x))^{2-1} . \cos (5x)^{1-1} . 5$$

Actually, \sin (x) = \cos (x). I simply forgot it while doing bigger equation. :D

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  • $\begingroup$ Looks like you did not really differentiate $\sin^2(5x)$ in the first one. And in the second one, you made a mistake when differentiating $x^3$ (which should differentiate to $3x^2$ and note $4x^2$). $\endgroup$
    – Kovomaka
    May 27 at 15:06
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    $\begingroup$ @Kovomaka Ohh! Sorry I did the mistake while typing.. In first term, I had differentiate using product rule. So, why I should differentiate $\sin^2$ at $3x^2 \sin^2 (5x)$? $\endgroup$
    – user876873
    May 27 at 15:07
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    $\begingroup$ Perhaps it helps to write $\sin^2(5x)$ as $\left(\sin(5x)\right)^2$. There are $3$ functions being applied to $x$: the multiply by $5$ function, the $\sin$ function, and the squaring function. $\endgroup$
    – Joe
    May 27 at 15:14
  • $\begingroup$ @Joe Is $\sin^2 (5x)=(\sin (5x))^2$ correct? $\endgroup$
    – user876873
    May 27 at 15:15
  • $\begingroup$ @Istiak: That's right. Some people have complained that this is an illogical notation—one would expect that $\sin^2\theta$ means $\sin(\sin(\theta))$—but this notation is so convenient that it's unlikely to be abandoned. $\endgroup$
    – Joe
    May 27 at 15:17
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Hint:

The derivative of the composition $\sin^2 5x$ is $$\frac{\mathrm d(\sin^2 5x)}{\mathrm dx}=\frac{\mathrm d(\sin^2 5x)}{\mathrm d(\sin 5x)}\cdot\frac{\mathrm d(\sin 5x)}{\mathrm d(5x)}\cdot\frac{\mathrm d(5x)}{\mathrm d x}=\dots$$

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  • $\begingroup$ I was trying to understand the hint. But, I couldn't get the value.. $$\frac{\mathrm d(\sin^2 5x)}{\mathrm dx}=\frac{\mathrm d(\sin^2 5x)}{\mathrm d(\sin 5x)}\cdot\frac{\mathrm d(\sin 5x)}{\mathrm d(5x)}\cdot\frac{\mathrm d(5x)}{\mathrm d x}=\dots$$ $\endgroup$
    – user876873
    May 27 at 15:14
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    $\begingroup$ It means you first take the derivative of $\sin^25x=(\sin 5x)^2$ as though $\sin 5x$ was the name of the variable, so that you get $2\sin 5x$, and so on. Is it clearer? $\endgroup$
    – Bernard
    May 27 at 15:18
  • $\begingroup$ Yes! It's clear now. I have tried again. Could you please see what am I missing now? $\endgroup$
    – user876873
    May 27 at 15:30
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    $\begingroup$ You final answer is correct. It can be further simplified to $3x^2 \sin^2 (5x) + 5 x^3 \sin 10x$. $\endgroup$
    – Bernard
    May 27 at 17:19
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$$\frac{dy}{dx}= \frac{d}{dx}(x^3)\times\sin^2(5x)+x^3\times \frac{d}{dx}(sin^2(5x))$$

$$\frac{dy}{dx} = 3x^2\sin^2(5x)+x^3 [ 2\sin(5x) ~\frac{dy}{dx}~( sin( 5x))]$$

$$\frac{dy}{dx} = 3x^2\sin^2(5x)+x^3 [ 2\sin(5x)\cos(5x) \frac{dy}{dx}(5x) ]$$

$$\frac{dy}{dx} = 3x^2\sin^2(5x)+x^3 [ 2\sin(5x)\cos(5x) ~5]$$

Note : $ \sin(2x) = 2 \sin x\cos x$

So, $5x^3 [ 2 \sin (5x) \cos (5x) = 5x^3 [sin(10x)= 2 \sin(5x) \cos(5x) ]$

$$\frac{dy}{dx} = 3x^2\sin^2(5x)+5x^3 \sin(10x)$$

You can cross verify the answer it should come the same

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  • $\begingroup$ The questions asks to use "Chain Rule" and not "Product Rule" $\endgroup$
    – Om3ga
    May 27 at 15:17
  • $\begingroup$ Could you please say how you wrote $$x^3 [2\sin (5x) \cos (5x) 5]=5x^3 10\sin (10x)$$? I have edited my question please check it also. $\endgroup$
    – user876873
    May 27 at 15:33
  • $\begingroup$ @Guarang: you have a typo/error at the very last step. $\endgroup$ May 27 at 15:34
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    $\begingroup$ @ancientmathematician I used double angle formula $\endgroup$
    – Gaurang
    May 27 at 15:40
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    $\begingroup$ You seem to equate $x^3[2\sin(5x)\cos(5x) 5]$ with $5x^3 10 \sin 10x$. $\endgroup$ May 27 at 15:50

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