$$ f(x)= an 11 x $$ the corresponding change in $$ y $$ in the form $$ d y=f^{\prime}(x) d x $$. $$ d y=\square d x $$

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$$ f(x)=	an 11 x $$ the corresponding change in $$ y $$ in the form $$ d y=f^{\prime}(x) d x $$. $$ d y=\square d x $$

UpStudy AI · Accepted Answer

To find the corresponding change in $$ y $$ in the form $$ dy = f'(x) dx $$ for the function $$ f(x) = \tan(11x) $$, we first need to compute the derivative $$ f'(x) $$.

The derivative of $$ \tan(kx) $$ is given by the formula:

$$
\frac{d}{dx} \tan(kx) = k \sec^2(kx)
$$

In our case, $$ k = 11 $$. Therefore, we can compute the derivative as follows:

$$
f'(x) = 11 \sec^2(11x)
$$

Now, we can express the change in $$ y $$ as:

$$
dy = f'(x) dx = 11 \sec^2(11x) \, dx
$$

Thus, the corresponding change in $$ y $$ in the form $$ dy = \square \, dx $$ is:

$$
dy = 11 \sec^2(11x) \, dx
$$

So, the answer is:

$$
dy = 11 \sec^2(11x) \, dx
$$
$$ dy = 11 \sec^2(11x) \, dx $$

\( f(x)=\tan 11 x \) the corresponding change in \( y \) in the form \( d y=f^{\prime}(x) d x \). \( d y=\square d x \)

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