Question

D
Differentiate
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Differentiate this

Answer 1

To find ---> Derivative of

xᵐ⁺ⁿ = aᵐ⁺ⁿ y²ⁿ

Solution---> We have some properties of log

(1) log ( m n ) = log m + log n

(2) log mⁿ = n log m

Now returning to original problem

xᵐ⁺ⁿ = aᵐ⁺ⁿ y²ⁿ

Taking log both sides we get

log ( xᵐ⁺ⁿ ) = log ( aᵐ⁺ⁿ y²ⁿ )

Applying above properties of log , we get

=> ( m + n ) logx = log aᵐ⁺ⁿ + log y²ⁿ

=> ( m + n ) logx = ( m + n ) loga + 2n logy

differentiating with respect to x , we get

=> (m+n) d/dx (logx) = d/dx {(m+n) loga}

+2n d/dx ( log y )

=> ( m + n ) ( 1 / x ) = 0 + 2n ( 1 / y ) (dy / dx)

=> ( m + n ) / x = ( 2n / y ) dy / dx

=> dy / dx = ( m + n ) y / 2nx

Additional information--->

(1) d / dx ( xⁿ ) = n xⁿ⁻¹

(2) d / dx ( eˣ ) = eˣ

(3) d / dx ( aˣ ) = aˣ loga

(4) d / dx ( logx ) = 1 / x

(5) d / dx ( Sinx ) = Cosx

(6) d / dx ( Cosx ) = - Sinx

(7) d / dx ( tanx ) = Sec²x

(8) d / dx ( Secx ) = Secx tanx

(9) d / dx ( Cotx ) = - Cosec²x

(10) d / dx ( Cosecx ) = - Cosecx Cotx

Answer 2

To find ---> Derivative of

xᵐ⁺ⁿ = aᵐ⁺ⁿ y²ⁿ

Solution---> We have some properties of log

(1) log ( m n ) = log m + log n

(2) log mⁿ = n log m

Now returning to original problem

xᵐ⁺ⁿ = aᵐ⁺ⁿ y²ⁿ

Taking log both sides we get

log ( xᵐ⁺ⁿ ) = log ( aᵐ⁺ⁿ y²ⁿ )

Applying above properties of log , we get

=> ( m + n ) logx = log aᵐ⁺ⁿ + log y²ⁿ

=> ( m + n ) logx = ( m + n ) loga + 2n logy

differentiating with respect to x , we get

=> (m+n) d/dx (logx) = d/dx {(m+n) loga}

+2n d/dx ( log y )