The differential for each function can be found:(a)y=x2sin⁡(4x)&nbsp;&nbsp;dy&nbsp;=&nbsp;(b)y=ln⁡((1+t2))&nbsp;&nbsp;dy&nbsp;=

b) Given function is y=ln⁡1+t2 Firstly, we apply rule of logarithms log⁡mn=nlog⁡m y=ln⁡1+t2 y=ln⁡(1+t2)12 y=12ln⁡(1+t2) Differential of log⁡x is dxx and that of x2 is 2xdx dy=121(2+t2)d(1+t2) dy=12(1+t2)2tdt dy=t(1+t2)dt

"The differential for each function can be found:(a)y=x2sin⁡(4x)dy=(b)y=ln⁡((1+t2))dy=" solved and explained

Fallbasiss4

Answered question

2022-01-23

The differential for each function can be found:
$(a) y = x^{2} \sin (4 x)$
$d y =$
$(b) y = \ln (\sqrt{(1 + t^{2})})$
$d y =$

Answer & Explanation

dodato0n

Beginner2022-01-24Added 9 answers

a) Given function is

y = x^{2} \sin 4 x

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

d y = (\sin 4 x) d (x^{2}) + x^{2} d (\sin 4 x)

Differential of

x^{2}

2 x d x

and that of

\sin 4 x

4 \cos 4 x

d y = (2 x \sin 4 x) d x + (4 x^{2} \cos 4 x) d x

d y = (2 x \sin 4 x + 4 x^{2} \cos 4 x)

Hana Larsen

Beginner2022-01-25Added 17 answers

b) Given function is

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

Firstly, we apply rule of logarithms

{\log m}^{n} = n \log m

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

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

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

Differential of

\log x

\frac{d x}{x}

and that of

x^{2}

2 x d x

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

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

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

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