Question

1. find \\( \frac { d y } { d x } \\) where \\( y = \frac { e ^ { x ^ { 2 } } \cos ^ { 2 } x } { x ^ { 2 } + 1 } + e ^ { x } \csc x \\).

Explanation:

Step1: Differentiate \(\frac{e^{x^{2}}\cos^{2}x}{x^{2}+1}\) using quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\)

Let \(u = e^{x^{2}}\cos^{2}x\), \(v=x^{2}+1\)
First, find \(u^\prime\) using product rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = e^{x^{2}}\), \(v=\cos^{2}x\)
\(u^\prime=e^{x^{2}}\cdot2x\cdot\cos^{2}x+e^{x^{2}}\cdot2\cos x(-\sin x)\)
\(v^\prime = 2x\)
Then \((\frac{u}{v})^\prime=\frac{(2xe^{x^{2}}\cos^{2}x - 2e^{x^{2}}\cos x\sin x)(x^{2}+1)-e^{x^{2}}\cos^{2}x\cdot2x}{(x^{2}+1)^{2}}\)

Step2: Differentiate \(e^{x}\csc x\) using product rule \((uv)^\prime = u^\prime v+uv^\prime\)

Let \(u = e^{x}\), \(v=\csc x\)
\(u^\prime=e^{x}\), \(v^\prime=-\csc x\cot x\)
\((e^{x}\csc x)^\prime=e^{x}\csc x - e^{x}\csc x\cot x\)

Step3: Combine the two derivatives

\(\frac{dy}{dx}=\frac{(2xe^{x^{2}}\cos^{2}x - 2e^{x^{2}}\cos x\sin x)(x^{2}+1)-2xe^{x^{2}}\cos^{2}x}{(x^{2}+1)^{2}}+e^{x}\csc x - e^{x}\csc x\cot x\)

Answer:

\(\frac{(2xe^{x^{2}}\cos^{2}x - 2e^{x^{2}}\cos x\sin x)(x^{2}+1)-2xe^{x^{2}}\cos^{2}x}{(x^{2}+1)^{2}}+e^{x}\csc x - e^{x}\csc x\cot x\)