Question
If $\text{y}=\text{x}^3\log\text{x},$ Prove that $\frac{\text{d}^4\text{y}}{\text{dx}^4}=\frac{6}{\text{x}}$
✓
Answer
here,
$\text{y}=\text{x}^3\log\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2\log{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log{\text{x}}+\text{x}^2$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\log\text{x}+6\text{x}\times\frac{1}{\text{x}}+5=6\log\text{x}+11$
Differentiating w.r.t.x, we get
$\text{y}=\text{x}^3\log\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2\log{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log{\text{x}}+\text{x}^2$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\log\text{x}+6\text{x}\times\frac{1}{\text{x}}+5=6\log\text{x}+11$
Differentiating w.r.t.x, we get
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