1 Find the gradient of the following curves at the point where $$ x=3 $$. a $$ y=\frac{x^{2}}{x-1} \quad $$ b $$ y=x \ln x \quad $$ c $$ f(x)=\frac{2 x^{2}-1}{x+3} \quad $$ d $$ s=\frac{t+5}{t^{2}+1} \quad $$ e $$ y=x^{2 x} $$

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**(a) For the curve** $$ y=\frac{x^2}{x-1} $$:

1. Write $$ u=x^2 $$ and $$ v=x-1 $$. Their derivatives are  
   $$
   u'=2x,\quad v'=1.
   $$
2. Apply the quotient rule:  
   $$
   y'=\frac{u'v-uv'}{v^2}=\frac{2x(x-1)-x^2}{(x-1)^2}.
   $$
3. Simplify the numerator:  
   $$
   2x(x-1)-x^2=2x^2-2x-x^2=x^2-2x=x(x-2).
   $$
4. So,  
   $$
   y'=\frac{x(x-2)}{(x-1)^2}.
   $$
5. At $$ x=3 $$:  
   $$
   y'(3)=\frac{3(3-2)}{(3-1)^2}=\frac{3\cdot1}{4}=\frac{3}{4}.
   $$

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**(b) For the curve** $$ y=x\ln x $$:

1. Use the product rule with $$ u=x $$ and $$ v=\ln x $$. Then,  
   $$
   u'=1,\quad v'=\frac{1}{x}.
   $$
2. The derivative is  
   $$
   y'=u'v+uv'= \ln x + x\left(\frac{1}{x}\right)=\ln x+1.
   $$
3. At $$ x=3 $$:  
   $$
   y'(3)=\ln3+1.
   $$

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**(c) For the curve** $$ f(x)=\frac{2x^2-1}{x+3} $$:

1. Let $$ u=2x^2-1 $$ (with $$ u'=4x $$) and $$ v=x+3 $$ (with $$ v'=1 $$).
2. The quotient rule gives:  
   $$
   f'(x)=\frac{4x(x+3)-(2x^2-1)}{(x+3)^2}.
   $$
3. Expand and simplify the numerator:  
   $$
   4x^2+12x-2x^2+1=2x^2+12x+1.
   $$
4. So,  
   $$
   f'(x)=\frac{2x^2+12x+1}{(x+3)^2}.
   $$
5. At $$ x=3 $$:  
   $$
   f'(3)=\frac{2(9)+12(3)+1}{(3+3)^2}=\frac{18+36+1}{36}=\frac{55}{36}.
   $$

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**(d) For the curve** $$ s=\frac{t+5}{t^2+1} $$:

1. Let $$ u=t+5 $$ (with $$ u'=1 $$) and $$ v=t^2+1 $$ (with $$ v'=2t $$).
2. Apply the quotient rule:  
   $$
   s'=\frac{1\cdot(t^2+1)-(t+5)(2t)}{(t^2+1)^2}.
   $$
3. Expand the numerator:  
   $$
   t^2+1-2t^2-10t=-t^2-10t+1.
   $$
4. So,  
   $$
   s'=\frac{-t^2-10t+1}{(t^2+1)^2}.
   $$
5. At $$ t=3 $$:  
   $$
   s'(3)=\frac{-(9)-10(3)+1}{(9+1)^2}=\frac{-9-30+1}{100}=\frac{-38}{100}=-\frac{19}{50}.
   $$

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**(e) For the curve** $$ y=x^{2x} $$:

1. Rewrite the function using the exponential form:  
   $$
   y=x^{2x}=e^{2x\ln x}.
   $$
2. Differentiate by recognizing that if $$ y=e^{g(x)} $$ then $$ y'=e^{g(x)}g'(x) $$. Here,  
   $$
   g(x)=2x\ln x.
   $$
3. Differentiate $$ g(x) $$ using the product rule:  
   $$
   g'(x)=2\ln x+2.
   $$
4. So,  
   $$
   y'=e^{2x\ln x}(2\ln x+2)=x^{2x}(2\ln x+2).
   $$
5. At $$ x=3 $$:  
   $$
   y'(3)=3^{6}\left(2\ln 3+2\right)=729\cdot2\left(\ln3+1\right)=1458\left(\ln3+1\right).
   $$
At $$ x=3 $$: - **a)** The gradient is $$ \frac{3}{4} $$. - **b)** The gradient is $$ \ln3 + 1 $$. - **c)** The gradient is $$ \frac{55}{36} $$. - **d)** The gradient is $$ -\frac{19}{50} $$. - **e)** The gradient is $$ 1458(\ln3 + 1) $$.

1 Find the gradient of the following curves at the point where \( x=3 \). a \( y=\frac{x^{2}}{x-1} \quad \) b \( y=x \ln x \quad \) c \( f(x)=\frac{2 x^{2}-1}{x+3} \quad \) d \( s=\frac{t+5}{t^{2}+1} \quad \) e \( y=x^{2 x} \)

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