1.

`t=(cos x)^(sin x+x^(2))` का x के सापेक्ष अवलन गुणांक ज्ञात कीजिये ।

Answer» `y=(cos x)^(sinx)+x^(x)`
`rArr y= u + v`
जहाँ `u=(cos x)^(sin x )`
और `v=x^(2)`
`:. (dy)/(dx)=(du)/(dx)+(dv)/(dx)` ...(1)
अब `u=(cos u ) ^(sin x )`
`rArr log u = sin x log cos x `
`rArr (1)/(u) (du)/(dx)= cos x log cos x + sin x (1)/( cos x) (-sin x)`
`rArr (1)/(u).(du)/(dx)=cos x log cos x - (sin ^(2)x)/ (cos x)`
`rArr (du)/(dx)=u { cos x log cos x -(sin^(2)x)/(cos x)}`
`rArr (du)/(dx)=(cos x )^(sin x) { cos x log cos x - (sin^(2)x)/(cos x)}` ...(2)
तथा `v=x^(2)`
`rArr log v = x log x`
`rArr (1)/(v) . (dv)/(dx)=x.(1)/(x)+log x .1`
`rArr (1)/(v). (dv)/(dx)=1+log x `
`rArr (dv)/(dx)=v(1+log x)`
`rArr (dv)/(dx)=x^(x)(1+log x)`
समीकरण (1 ) , (2 ) व (3 ) से,
`(dy)/(dx)=(cos x)^(sin x){cos x log cos x - (sin^(2)x)/(cos x)}+x^(x)(1+log x)`


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