Fallbasiss4

Answered

2022-01-23

The differential for each function can be found:

$\left(a\right)y={x}^{2}\mathrm{sin}\left(4x\right)$

$dy=$

$\left(b\right)y=\mathrm{ln}\left(\sqrt{(1+{t}^{2})}\right)$

$dy=$

Answer & Explanation

dodato0n

Expert

2022-01-24Added 9 answers

a) Given function is

$y={x}^{2}\mathrm{sin}4x$

Here, we use product rule of differentiation to find the differential

$dy=\left(\mathrm{sin}4x\right)d\left({x}^{2}\right)+{x}^{2}d\left(\mathrm{sin}4x\right)$

Differential of$x}^{2$ is $2xdx$ and that of $\mathrm{sin}4x$ is $4\mathrm{cos}4x$

$dy=\left(2x\mathrm{sin}4x\right)dx+\left(4{x}^{2}\mathrm{cos}4x\right)dx$

$dy=(2x\mathrm{sin}4x+4{x}^{2}\mathrm{cos}4x)$

Here, we use product rule of differentiation to find the differential

Differential of

Hana Larsen

Expert

2022-01-25Added 17 answers

b) Given function is

$y=\mathrm{ln}\sqrt{1+{t}^{2}}$

Firstly, we apply rule of logarithms${\mathrm{log}m}^{n}=n\mathrm{log}m$

$y=\mathrm{ln}\sqrt{1+{t}^{2}}$

$y={\mathrm{ln}(1+{t}^{2})}^{\frac{1}{2}}$

$y=\frac{1}{2}\mathrm{ln}(1+{t}^{2})$

Differential of$\mathrm{log}x$ is $\frac{dx}{x}$ and that of $x}^{2$ is $2xdx$

$dy=\frac{1}{2}\frac{1}{(2+{t}^{2})}d(1+{t}^{2})$

$dy=\frac{1}{2(1+{t}^{2})}2tdt$

$dy=\frac{t}{(1+{t}^{2})}dt$

Firstly, we apply rule of logarithms

Differential of

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