Question

1. find $\frac{dy}{dx}$ where $y=\frac{e^{x^{2}}}{cos^{2}x}+e^{x}csc x + x^{2}+1$

Explanation:

Step1: Apply quotient rule for $\frac{e^{x^{2}}}{\cos^{2}x}$

Let \(u = e^{x^{2}}\), \(v=\cos^{2}x\). Then \(u^\prime=e^{x^{2}}\cdot2x\), \(v^\prime = 2\cos x(-\sin x)\). By quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), we have \(\frac{d}{dx}(\frac{e^{x^{2}}}{\cos^{2}x})=\frac{2xe^{x^{2}}\cos^{2}x+2e^{x^{2}}\sin x\cos x}{\cos^{4}x}\).

Step2: Differentiate \(\frac{1}{x^{2}+1}\)

Using the power rule \((x^{n})^\prime = nx^{n - 1}\) and quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) with \(u = 1\), \(v=x^{2}+1\), we get \(\frac{d}{dx}(\frac{1}{x^{2}+1})=\frac{0\times(x^{2}+1)-1\times2x}{(x^{2}+1)^{2}}=-\frac{2x}{(x^{2}+1)^{2}}\).

Step3: Differentiate \(e^{x}\csc x\)

Using product rule \((uv)^\prime=u^\prime v+uv^\prime\) with \(u = e^{x}\), \(v=\csc x\). Since \(u^\prime=e^{x}\), \(v^\prime=-\csc x\cot x\), we have \(\frac{d}{dx}(e^{x}\csc x)=e^{x}\csc x - e^{x}\csc x\cot x\).

Step4: Sum up the derivatives

\(\frac{dy}{dx}=\frac{2xe^{x^{2}}\cos^{2}x + 2e^{x^{2}}\sin x\cos x}{\cos^{4}x}-\frac{2x}{(x^{2}+1)^{2}}+e^{x}\csc x - e^{x}\csc x\cot x\)

Answer:

\(\frac{2xe^{x^{2}}\cos^{2}x + 2e^{x^{2}}\sin x\cos x}{\cos^{4}x}-\frac{2x}{(x^{2}+1)^{2}}+e^{x}\csc x - e^{x}\csc x\cot x\)