Question

let $y = \tan(3x + 5)$. find the differential $dy$ when $x = 1$ and $dx = 0.2$ find the differential $dy$ when $x = 1$ and $dx = 0.4$

Explanation:

Step1: Find derivative of y

The function is \( y = \tan(3x + 5) \). Using the chain rule, the derivative \( y'=\frac{dy}{dx}=3\sec^{2}(3x + 5) \). So, \( dy = 3\sec^{2}(3x + 5)dx \) (here \( dx = dz \) as per notation in the problem).

Step2: Evaluate at x = 1

First, calculate \( 3x+5 \) when \( x = 1 \): \( 3(1)+5 = 8 \). Then, \( \sec^{2}(8)=\frac{1}{\cos^{2}(8)} \). But we can also use the identity \( \sec^{2}\theta=1 + \tan^{2}\theta \), but directly, we know that \( \frac{dy}{dx}=3\sec^{2}(3x + 5) \), so at \( x = 1 \), \( \frac{dy}{dx}=3\sec^{2}(8) \). However, we can also note that \( \sec^{2}(8)=1+\tan^{2}(8) \), but maybe it's easier to just compute the value. Wait, actually, let's recall that \( \frac{d}{dx}\tan(u)=u'\sec^{2}(u) \), so for \( u = 3x + 5 \), \( u' = 3 \), so \( y' = 3\sec^{2}(3x + 5) \). So \( dy=3\sec^{2}(3x + 5)dx \). Now, when \( x = 1 \), \( 3x + 5=8 \), so \( dy = 3\sec^{2}(8)dx \). Now, \( \sec^{2}(8)=\frac{1}{\cos^{2}(8)} \). Let's compute \( \cos(8) \) (assuming \( x \) is in radians? Wait, the problem doesn't specify, but in calculus, we usually use radians. So \( \cos(8)\approx\cos(8 - 2\pi)\approx\cos(8 - 6.283)\approx\cos(1.717)\approx0.185 \), so \( \cos^{2}(8)\approx0.0342 \), so \( \sec^{2}(8)\approx\frac{1}{0.0342}\approx29.24 \). Then \( 3\sec^{2}(8)\approx3\times29.24 = 87.72 \). Wait, but maybe there's a mistake here. Wait, no, wait: \( 8 \) radians is about \( 8\times\frac{180}{\pi}\approx458 \) degrees, which is \( 458 - 360 = 98 \) degrees, so \( \cos(8)=\cos(98^\circ)\approx - 0.139 \), so \( \cos^{2}(8)\approx0.0193 \), so \( \sec^{2}(8)\approx\frac{1}{0.0193}\approx51.81 \), then \( 3\sec^{2}(8)\approx155.43 \). Wait, I think I miscalculated the reduction. \( 8 \) radians: \( 2\pi\approx6.283 \), so \( 8 - 2\pi\approx1.717 \) radians, which is about \( 1.717\times\frac{180}{\pi}\approx98.37 \) degrees, so \( \cos(1.717)\approx\cos(98.37^\circ)\approx - 0.145 \), so \( \cos^{2}(1.717)\approx0.0210 \), so \( \sec^{2}(1.717)\approx47.62 \), so \( 3\times47.62 = 142.86 \). Hmm, maybe I should use a calculator for better precision. Let's use a calculator: \( \cos(8)\approx\cos(8)\approx - 0.1455000339 \), so \( \cos^{2}(8)\approx0.021170309 \), so \( \sec^{2}(8)\approx\frac{1}{0.021170309}\approx47.236 \), so \( 3\sec^{2}(8)\approx3\times47.236 = 141.708 \). So \( dy = 141.708\times dx \).

Now, first case: \( dx = 0.2 \)

Step3: Compute dy for dx=0.2

\( dy = 3\sec^{2}(8)\times0.2 \). Since \( 3\sec^{2}(8)\approx141.708 \), then \( dy\approx141.708\times0.2 = 28.3416 \)

Second case: \( dx = 0.4 \)

Step4: Compute dy for dx=0.4

\( dy = 3\sec^{2}(8)\times0.4\approx141.708\times0.4 = 56.6832 \)

Wait, but maybe there's a better way. Wait, actually, the problem is about differentials, so \( dy = y'(x)dx \). So first, find \( y'(x)=3\sec^{2}(3x + 5) \). At \( x = 1 \), \( 3x + 5 = 8 \), so \( y'(1)=3\sec^{2}(8) \). Now, \( \sec^{2}(8)=1+\tan^{2}(8) \). Let's compute \( \tan(8) \). \( \tan(8)=\tan(8 - 2\pi)=\tan(1.717)\approx\tan(98.37^\circ)=-\tan(8.37^\circ)\approx - 0.146 \)? Wait, no, \( \tan(98.37^\circ)=\tan(90^\circ + 8.37^\circ)=-\cot(8.37^\circ)\approx - \frac{1}{\tan(8.37^\circ)}\approx - \frac{1}{0.146}\approx - 6.849 \). So \( \tan^{2}(8)\approx( - 6.849)^{2}\approx46.91 \), so \( \sec^{2}(8)=1 + 46.91 = 47.91 \), so \( 3\sec^{2}(8)=3\times47.91 = 143.73 \). Then, for \( dx = 0.2 \), \( dy = 143.73\times0.2 = 28.746 \). For \( dx = 0.4 \), \( dy = 143.73\times0.4 = 57.492 \). The slight difference is due to approximation of \( \tan(8) \) or \( \cos(8) \). Alternatively, using a calculator for more precision:

Using a calculator, \( \cos(8)\approx - 0.1455000339 \), so \( \sec^{2}(8)=\frac{1}{\cos^{2}(8)}\approx\frac{1}{0.021170309}\approx47.236 \), so \( 3\sec^{2}(8)\approx141.708 \). Then:

First dy (dx=0.2): \( 141.708\times0.2 = 28.3416 \)

Second dy (dx=0.4): \( 141.708\times0.4 = 56.6832 \)

So the approximate values are 28.34 (when dx=0.2) and 56.68 (when dx=0.4). But let's check with exact steps:

1. Find the derivative of \( y = \tan(3x + 5) \):

Using the chain rule, if \( y = \tan(u) \) and \( u = 3x + 5 \), then \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\sec^{2}(u)\cdot3 = 3\sec^{2}(3x + 5) \).

2. The differential \( dy \) is given by \( dy=\frac{dy}{dx}dx = 3\sec^{2}(3x + 5)dx \).

3. Substitute \( x = 1 \) and \( dx = 0.2 \):

First, compute \( 3x + 5 \) when \( x = 1 \): \( 3(1)+5 = 8 \).

So \( dy = 3\sec^{2}(8)\cdot0.2 \).

Now, \( \sec^{2}(8)=\frac{1}{\cos^{2}(8)} \). Using a calculator, \( \cos(8)\approx - 0.1455 \), so \( \cos^{2}(8)\approx0.02117 \), so \( \sec^{2}(8)\approx47.236 \).

Thus, \( dy = 3\times47.236\times0.2 = 141.708\times0.2 = 28.3416 \approx28.34 \).

4. Substitute \( x = 1 \) and \( dx = 0.4 \):

\( dy = 3\sec^{2}(8)\cdot0.4 = 141.708\times0.4 = 56.6832 \approx56.68 \).

Answer:

When \( x = 1 \) and \( dx = 0.2 \), \( dy\approx28.34 \) (or more precisely, using calculator values, maybe more accurate). When \( x = 1 \) and \( dx = 0.4 \), \( dy\approx56.68 \).

Wait, but maybe there's a mistake in assuming radians? Wait, the problem doesn't specify, but in calculus, we always use radians for differentiation. So the calculations are correct. Alternatively, if we use degrees, but that would be wrong because the derivative formulas are for radians. So the answer should be approximately 28.34 (for dx=0.2) and 56.68 (for dx=0.4).