By using the properties of differentiation of the function of power " n ", where n is constant, we can answer the following question
1) y = ( x + x + x ) y ′ = 1 2 ( x + x + x ) ∗ ( 1 + 1 + 1 2 x 2 x + x ) \begin{aligned}
&\text { 1)} y=\sqrt{(x+\sqrt{x+\sqrt{x}})}\\
&y^{\prime}=\frac{1}{2 \sqrt{(x+\sqrt{x+\sqrt{x}})}} *\left(1+\frac{1+\frac{1}{2 \sqrt{x}}}{2 \sqrt{x+\sqrt{x}}}\right)
\end{aligned} 1) y = ( x + x + x ) y ′ = 2 ( x + x + x ) 1 ∗ ( 1 + 2 x + x 1 + 2 x 1 )
By using the properties of differentiation of the exponential functions, we get the answer of the following question
2 ) y = e cos x + cos e x y ′ = ( − sin x ) ∗ e cos x − e x ( sin e x ) \begin{aligned}
&2)\,y=e^{\cos x}+\cos e^{x}\\
&y^{\prime}=(-\sin x) * e^{\cos x}-e^{x}\left(\sin e^{x}\right)
\end{aligned} 2 ) y = e c o s x + cos e x y ′ = ( − sin x ) ∗ e c o s x − e x ( sin e x )
Using the properties of differentiation of the Logarithmic functions, we get the answer of the following question
3 ) y = ln ( x ln x ) y ′ = ln x + x ∗ 1 x x ln x = ln x + 1 x ln x \begin{aligned}
&3)y=\ln (x \ln x)\\
&y^{\prime}=\frac{\ln x+x * \frac{1}{x}}{x \ln x}=\frac{\ln x+1}{x \ln x}
\end{aligned} 3 ) y = ln ( x ln x ) y ′ = x ln x ln x + x ∗ x 1 = x ln x ln x + 1
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